KEMA JOINT, MATHEMATICS PAPER 1 SECTION 1

Added: 2026-05-20

[00:00:02]Thank you so much for joining this lesson.

[00:00:05]We're going to discuss the Kisumu West Mathematics Association, that is the Kima mathematics paper one.

[00:00:13]This is a very serious joint examination paper.

[00:00:17]And it's my hope that through the discussion you guys are going to get something. So, we will start the paper with section one.

[00:00:27]Number one.

[00:00:29]We are told to evaluate 0.04 minus 0.

[00:00:35]0018 divided by 0.06 plus 0.002 over half of 0.048.

[00:00:50]So, we can start with simplifying the numerator first.

[00:00:56]Whereby according to BODMAS you know this is the acronym that will help us to know how signs are supposed to be ordered. For example, we start with the brackets and they are not there.

[00:01:12]Of is not there, but division is there.

[00:01:15]So, because there's division, we start with it. So, it will be -0.0018 divided by 0.06.

[00:01:26]This will give us -0.03.

[00:01:30]So, now we are having 0.04 -0.03 +0.00 2. So, we what you're having now is minus and plus which can be done at the same time now.

[00:01:51]So, 0.

[00:01:53]04 -0.03 then we add 0.00 2.

[00:02:02]Mhm.

[00:02:03]I'm getting 0.012.

[00:02:08]That is now from the numerator. What about the denominator?

[00:02:14]Whereby at the denominator, it's only off which is there.

[00:02:19]And with that, we supposed to multiply 1/8 multiplied by 0.0 48.

[00:02:29]So, this shall be Mhm.

[00:02:34]0.

[00:02:36]0.048 multiplied by 1/8.

[00:02:40]This shall be 0.

[00:02:47]0.

[00:02:48]0.006.

[00:02:51]Like that. 0.006.

[00:02:54]So, we're having now our simplified part of the numerator.

[00:02:58]Simplified part of the denominator.

[00:03:01]And now we ought to combine.

[00:03:03]Remember we are having now 0.012 out of 0.006.

[00:03:15]So, at this point now, can we talk about doing away with the decimals by multiplying by 1,000 in each side? So, 12 out of six giving us a simplified answer as exactly two.

[00:03:30]So, we supposed to work out step-by-step until you get the final answer, which is positive two.

[00:03:38]In number one.

[00:03:39]So, we check number two now. The question in number two says that uh the size of each interior angle of a regular polygon is seven times the size of the exterior angle. Find the number of sides of the polygon.

[00:03:55]One thing that you're supposed to know is that the interior and exterior angles of a polygon are usually related, and the relation is that when we take interior angle, we add the exterior angle, we're supposed to get 180.

[00:04:12]Uh-huh.

[00:04:13]And then you can see there is a relation being given here. Each interior is seven times the exterior. So, if exterior is let to be X, then interior shall automatically become seven times the value of exterior.

[00:04:34]And now from this relation, we can equate now X plus seven X should give us 180, meaning that 8 X equals to 180, meaning that the value of X equals to 180 out of eight.

[00:04:57]This gives us 22.5.

[00:05:02]22.5. So, the size of the exterior angle, which is the value of X, is 22.5.

[00:05:13]Then now, we see that uh the sum of exterior angles usually gives 360.

[00:05:22]So, we can get the number of exterior angles by saying that uh each exterior angle is 22.5.

[00:05:32]When you add all of them, you multiply by their number, which is n.

[00:05:36]They should total to 360.

[00:05:40]Therefore, to remain with n, you've now to divide both sides by 22.5. So, 360 out of 22.5 and we get exactly 16.

[00:05:53]So, the number of sides is 16.

[00:05:56]Because angles in a polygon equal to the number of sides. So, if we are getting 16 exterior angles, it means we have 16 sides making up the polygon. A very good question. If you're getting something remember to comment on the comment section of this video that I am learning. You can also consider joining our programs. If you are a day scholar, please join Shifting Grades Online School. We have very serious programs for day scholars during term two and also term three.

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[00:08:17]Number three in this mathematics paper, solve for x in the equation 2 ^ 1 - x - 2 ^ x + 2 = 7. Allow me apply the first and the second laws of indices whereby we say when we have a ^ m multiplied by a ^ n you simply have to take a raised to the power of m + n.

[00:08:46]Then when we have a ^ m divided by a ^ n then we supposed to take uh the value of a which is the base, then the powers are subtracted. So, with these two laws then actually we call it the multiplication law and the division law of indices. So, we can apply those laws at the first term and also the second one. Whereby we shall say this is a You see a minus on the powers, then it means we have 2 ^ 1 divided by 2 ^ x. Then minus 2 ^ x multiplied by 2 ^ 2 because this is addition.

[00:09:34]So, it has been multiplied by 2 power two, which gives us four.

[00:09:40]This is exactly equal to seven. But now you can see uh two power x is appearing in two cases, at this point and also at this point. So can we let two power x to be a different unknown. We can even let it to be t. Then we proceed by saying this is two out of t minus uh-huh two by t cuz uh Okay, four by t because two power two equals to four. So four times two power x, which is our t. This is equal to seven. Can we now multiply throughout by the value of t such that two remains the way it is, then minus four t gets squared equals to seven. We introduce t at that point. So I've multiplied all the terms by t to do away with the fraction, or rather to do away with the denominator at that particular point.

[00:10:46]So with that we can now arrange the equation such that it becomes quadratic. Then we proceed to solve it. Uh-huh, 4t squared can be on the other side becoming positive. Then on that same side there is already 7t.

[00:11:07]But two becomes positive on that other side. So that we remain with a zero on the same side.

[00:11:14]Yeah, zero on the same side. So we can use the factorization method, though methods of solving quadratic equations are very many. Allow me use the factorization method.

[00:11:27]Whereby you can say we need a product a product of a times c. Four by -2 giving us -8.

[00:11:35]Then sum of 7.

[00:11:39]Two terms when multiplied they give us -8 and when added they give us a positive 7.

[00:11:50]When we check 8 and -1 you discover that the product is -8, the sum is positive 7. So now we can replace 7t with the two factors identified.

[00:12:05]4 t squared plus 8t -t.

[00:12:12]That is now 7 split into 8 and -1. Then -2 equals to 0.

[00:12:19]From the first two factors we can factor out 4t, which is common.

[00:12:25]We remain with t plus 2.

[00:12:28]Then -1. So that we remain again with t plus 2.

[00:12:33]This is equal to 0. With this now we can say from inside the brackets we have a t plus 2 and outside the brackets we have 4t -1.

[00:12:47]This is equal to 0.

[00:12:50]Uh-huh.

[00:12:52]If the product equals to 0, then any of the factors can be 0 or both of them. So t plus 2 equals to 0.

[00:13:03]equals to 0. Meaning that the value of t is negative 2. Or 4t -1 equals to 0. Meaning that 4t equals to positive 1 and t becomes a quarter.

[00:13:20]So we have two possible values of t.

[00:13:23]Two possible values of t.

[00:13:26]With the two possible values of t with two possible values of t we can now check what we had let to be t.

[00:13:37]What had we let to be t?

[00:13:39]Uh-huh.

[00:13:40]It is 2 ^ x which had been let to be t.

[00:13:45]And this is equal to The first option we can use a quarter.

[00:13:51]So that we say 2 ^ x equals to half squared.

[00:14:01]At this point we can apply the law of indices that says with a fraction like this or with a reciprocal like this, we can talk of 2 ^ x equals to instead of half, we take the reciprocal to get two, then the power also changes to a positive two.

[00:14:21]To a negative two, that is.

[00:14:22]So at this point, ladies and gentlemen, you can now conclude with me that if 2 ^ x equals to 2 ^ -2 then x equals to negative two.

[00:14:35]So the value of x is negative two. On the other side, you should know that uh when we start using a negative base then that is a very high level of math.

[00:14:47]So we can now ignore this value of t and we operate with t equals to a quarter because at this level we only deal with positive bases.

[00:15:00]Yeah, and also logarithms we deal with positive logarithms.

[00:15:05]So we now proceed with that part.

[00:15:09]Uh-huh. Can we check number four of this paper? Yes.

[00:15:14]So now to Now to number four.

[00:15:18]We are told that we simplify the expression x out of x minus y uh minus y out of x plus y minus 2xy out of x squared minus y squared.

[00:15:34]One thing that we're supposed to know is that we're supposed to use what you call the LCM to simplify.

[00:15:41]The LCM here you see this is a difference of two squares at this point. That's a difference of two squares. Therefore, do you remember that when you have a perfect square minus another perfect square, we call it a difference of two squares and it's usually simplified to a minus b a plus b. So, the simplest form of a difference of two squares is this one. I'm giving a general a general simplification of a difference of two squares according to quadratic identities. That is a book to work.

[00:16:25]So, now we can see that this one is a factor in this.

[00:16:29]This one, eh?

[00:16:31]The first denominator is a factor in the last one. And also the second denominator is a factor in the last one.

[00:16:37]So, being factors all of them being factors then we can say that the square root is x minus one x plus y. So, with this understanding do you see now that we can pick the LCM we divide with the first denominator and after the division it is the positive part that shall remain. So, we multiply now by the numerator. So, it will be x times x plus y because it is the negative part that are simplified. When you pick again the LCM, you divide by the second denominator.

[00:17:20]The one with positive cancels out and you remain with minus y into x minus y.

[00:17:28]Then minus, uh-huh, we divide the LCM by this one and the whole of it will cancel because it is just the same thing only that on the LCM I have simplified. And therefore, after they have canceled out one another, we remain with one. And that is why 2 xy shall remain like that.

[00:17:51]At this point now, we can open brackets and we also group terms together. For example, x becomes squared minus xy. Sorry, that is a positive.

[00:18:04]Then here we add xy uh-huh, positive y squared.

[00:18:10]Then on this part, we have a negative 2 xy.

[00:18:16]Then everything is being divided by Mhm.

[00:18:22]x minus y into x plus y.

[00:18:29]Are we together at that point? If yes, type yes.

[00:18:34]Then now at this point, it is good I remind you that you're supposed to be very careful. I said x by x, it became squared. x by y, xy. But at this point, y by x, there is a negative.

[00:18:47]So it's negative xy.

[00:18:49]Negative y and negative y, it is y squared. So that is what I wanted to remind you. Mhm. You can see now, this and this one will cancel out. Then now on the numerator, we can start with x squared. Want to make it a quadratic.

[00:19:06]Uh-huh. So negative 2x can follow. 2xy can follow there. Then we finish with y squared. You look at that.

[00:19:15]Then, everything out of x - y x + y Uh-huh. That is a properly factorized.

[00:19:31]Now, at this point, allow me work on the numerator, which is a difference of two square Oh, no. It's a quadratic expression, actually. So, we can talk about a product of one then sum being -2. Because this is 1 by 1, but at the middle, we have -2.

[00:19:50]So, this is -1 and -1. Multiplied, positive 1. Added, -2.

[00:19:58]So, now we have x squared xy xy once again then + y squared. I'm dealing with the numerator first. Uh-huh. This is a Okay. Or I can have divide by Yeah, we can carry on with a denominator throughout.

[00:20:20]So, x - y x + y At this point from the first two terms, we can factor out x. Then, we remain with x - y. In the second, we factor out a -y.

[00:20:39]We remain with x - y.

[00:20:43]Uh-huh.

[00:20:44]And remember everything is out of what?

[00:20:48]Everything is out of x - y x + y. So, from the numerator inside the brackets x - y outside the brackets x - y again. This is a what you call a perfect square.

[00:21:09]Out of x - y x + y. So, with this now, you can see x - y x - y can cancel out.

[00:21:24]Then we remain with x - y out of x + y.

[00:21:31]So, that is now the final answer and the most simplified part of this particular expression.

[00:21:41]We get to number five now. Yeah, we get to number five. Now, three bells rings at intervals of 24, 30, and 36 minutes.

[00:21:51]They all ring together at 10:00 a.m. Determine the number of times the 24 minutes bell will ring before they all ring together. So, can we check the time required for them to ring together?

[00:22:10]Uh-huh. That is going to be the interval. So, the interval required for them to ring together again shall be the LCM of 24, 30, and 36.

[00:22:24]LCM of 24, 30, 36. So, we can check out that.

[00:22:30]Let us check out that.

[00:22:33]Let us check out that.

[00:22:35]Uh-huh.

[00:22:36]We divide by two everything, get 12.

[00:22:41]Get 15.

[00:22:42]We get 18.

[00:22:43]Then from this by two uh we get six.

[00:22:49]15 remains. This one comes nine.

[00:22:52]By two once again, three, 15, nine. Now, we can go to three, one, five, three.

[00:23:02]So, by 3 once again, 1 5 1.

[00:23:07]And by 5, everything now becomes a 1.

[00:23:11]Meaning that the least common multiple is 2 * 2 * 2 * 3 * 3 * 5.

[00:23:25]So, we can work out that 2 ^ 3 3 ^ * 5 8 * 9 * 5 I'm getting 360 minutes.

[00:23:40]So, this is the interval they need for them to ring together.

[00:23:45]Now, after getting the number of minutes required and then we know very well that the 24 one of the bells is ringing after 24.

[00:23:55]How many times will it ring?

[00:23:57]360 out of 24. So, how many intervals will be here?

[00:24:03]We divide by 24 and I'm getting exactly 15 times.

[00:24:10]Yeah.

[00:24:12]15 times. So, that is the number of times we shall have the 24 minutes bell ring.

[00:24:21]A very good question on the application of LCM.

[00:24:26]So, we can get number six now. And guys, uh remember to subscribe to the best education platform and that is Shifting Grades Online School. So, you're most welcome uh to this great platform. Remember to just subscribe. I told you the other time that we are now at 90 2,000 subscribers and we are added to the Silver Creator Award, which is given at 100,000 subscribers. I invite you to be my guest. Please, don't be left out.

[00:25:00]Kindly be a subscriber. Stand to be counted.

[00:25:05]Now, form all the inequalities that satisfy the region R in the figure below. So, inequalities have already been plotted, three of them. L1, L2, L3, L4. Okay. 1 2 3 three of them.

[00:25:22]They are shielding region R. So, which inequalities are these?

[00:25:27]First, you're supposed to know that we come up with we're supposed to come up with a equations first.

[00:25:37]Y equals to mx + c. This is the best format that you should have your equations at.

[00:25:44]Uh-huh. We know very well that what you have here is the gradient.

[00:25:50]We know very well that what you have here is um the Y intercept. Y intercept is the point where the line is cutting the Y axis. Now that we can interpret gradient and also Y intercept, then coming up with the lines can become very easy. For example, the first line L1, this one is not having um this one is not having any gradient because it's an horizontal line.

[00:26:22]But now that it doesn't have a gradient, it has a Y intercept which is a four.

[00:26:28]Therefore, that is line Y equals to with no gradient, it means m is a zero times x, getting a zero. So, Y equals to four, that is the first line.

[00:26:39]Then, with a line like that, we can now get the inequality. According to the shaded part of the line, those are values greater than Y, but the region R is on the lower side where values of Y are less than four.

[00:26:54]Therefore, it means Y is less than four.

[00:26:58]And another factor is that L1 is a complete line. When it is a complete line, I mean not dotted.

[00:27:06]When it is a complete line then we add or equal to. So, this is the borderline for line one, that is the inequality.

[00:27:17]What about line two? Line two is here.

[00:27:19]So, let us check line two. You can see it is a dotted line.

[00:27:22]Uh for you to get the gradient uh before you even proceed, it's a positive gradient according to how the line is sloping. Therefore, you can pick two fixed coordinates like that one and this one.

[00:27:36]Then you check the triangle that they make this way.

[00:27:40]Check the triangle that they make this way.

[00:27:44]Then we know very well gradient for the line two is usually the change in Y over change in X.

[00:27:53]So, from this point to this one there are three units.

[00:27:58]And horizontally from this point to this and we are considering the right angle there.

[00:28:03]We are having two units. So, the gradient is three over two. Meaning that now the equation shall be Y equals to gradient three out of two X plus where is this line intersecting? Where is this line intersecting the vertical axis at negative two?

[00:28:25]At negative two. So, at negative two now instead of a plus we shall have a negative two at that point. So, we have the equation for L2 because we have fixed the gradient where it should be. We have also fixed the Y intercept where it should be. Now, what remains is to find the correct symbol of this inequality.

[00:28:46]For that we can pick a point which is on the required side.

[00:28:50]Like a 3 3. Yeah, 0.33.

[00:28:54]If we pick 0.33, can we substitute where there is Y 3, where there is X 3?

[00:29:00]So, Y is a 3.

[00:29:03]Uh-huh, where there is X 3 over 2 by 3 you get 9 over 2.

[00:29:07]Then minus 2.

[00:29:08]So, 9 out of 2 minus 2, I'm getting 2 and 1/2. So, 3 and 2 and 1/2. 3 Then this is 2 and 1/2.

[00:29:18]3 is greater than 2 and 1/2.

[00:29:21]So, now this is the suitable inequality.

[00:29:24]So, Y is greater than 3 out of 2 X minus 2.

[00:29:30]This is line 2.

[00:29:34]Line 2 like that.

[00:29:37]So, we have our second line there.

[00:29:40]We can now proceed using the same concept to line 3. For line 3, we can see there is no Y intercept or rather Y intercept is a zero. So, we shall be adding zero where there is a constant because it is passing through the origin. Then Now, for the gradient of the third line, for the gradient of the third line, we can also check fixed coordinates for this particular line.

[00:30:06]Fixed coordinates. So, let's get to the fixed coordinates.

[00:30:10]Uh-huh, we can use um even from this point to this one, yeah?

[00:30:19]Within those two points which are fixed, can see the number of units on the vertical, number of units on the horizontal.

[00:30:29]So, on the vertical you can see from this point to this one, only two units.

[00:30:33]Then here one unit. So, it's like two over one, which is simply two. Therefore, the equation shall be Y equals to gradient of two times x plus a y intercept of zero. So, it remain like that. We can use the same same coordinate which is a 3 3 because it is within the required region. So, we have a three where there is y and where there is x also three. So, three equals to two by three which is a six.

[00:31:03]And three is less than six. Meaning that this inequality now shall be less than and because line three because line three is a complete line we introduce or equal to 2x. So, for line three this is the inequality.

[00:31:25]For line three we have the inequality. So, that is how you come up with the inequalities in a very quick way without having to waste your time. Very easy.

[00:31:37]Very easy. If you're getting something kindly type on our comment section that yes, I am learning.

[00:31:44]Number seven, ladies and gentlemen.

[00:31:47]The masses of two similar solid cylinders are 4752 g and 1408 g.

[00:31:56]Uh-huh.

[00:31:57]If the difference in area mark that one very quickly.

[00:32:02]The difference in area of the curved surfaces of the two cylinders is 280 find the area of the curved surface of the larger cylinder. The first thing that you're supposed to know in this question is that mass is three-dimensional. Therefore, ratios of the masses will give us what we call what we call a volume scale factor.

[00:32:32]Volume scale factor.

[00:32:35]This is a 47 52 1408.

[00:32:43]Can this be simplified?

[00:32:47]4752 and 1408.

[00:32:54]Yeah, this simplifies to Actually, if I write it as a fraction, 3 and 3 over 8.

[00:33:02]You see 3 and 3 over 8 means uh means 27 over 8.

[00:33:09]27 over 8 or if it must be a ratio, 27 is to 8. So, the ratio of the two volumes is 27 is to 8, meaning that with the ratios of the volumes, we can now find the the relation between any linear measurements. We call it linear scale factor.

[00:33:31]Whereby, it is the cube the cube root of the cube root of volume scale factor gives us linear scale factor. This shall be 3 is to 2.

[00:33:46]Then, now area scale factor because at the end, I can see areas have been related. So, if areas have been related, then we have to get to area scale factor. Whereby, you square.

[00:34:03]This shall be 9 is to 4.

[00:34:07]Therefore, area of the larger is to area of the smaller 9 is to 4. If these are the ratios, then can we get the difference in the ratios?

[00:34:19]The difference in the ratios is 5. So, 5 proportions the difference in the areas.

[00:34:25]Five proportions are the difference, and this is corresponding to 280.

[00:34:33]Meaning that area of the larger which is now nine proportions shall be 280 by nine over five.

[00:34:45]280 by nine over five. This is giving us 504.

[00:34:54]504 square centimeters. So, that is the area of the larger.

[00:35:00]Very good question.

[00:35:02]You can also say that uh 280 corresponds to five. What about the smaller one which is four proportions such that you have 280 by four over five and it will give you 224 square centimeters. So, when you relate 504 and 204 224 you will get the same same ratio of nine is to four.

[00:35:31]So, you will confirm that this answer indeed is the right one.

[00:35:36]This answer 504 is the right answer because when you proceed to the smaller one, then you compare the ratios as five over four and 224 they will simplify to nine is to four.

[00:35:50]Meaning that these are the correct values, but we simply wanted for the larger.

[00:35:55]A very good question. What you should master is that masses densities, volumes, and capacities are three-dimensional.

[00:36:05]They are three-dimensional uh quantities and they give us volume scale factor.

[00:36:13]Proceeding to number eight which is a question on scale drawings.

[00:36:18]The boundaries AB, BC, CD, and DA of a ranch are straight lines such that B is 12 km on a bearing of 045 from A.

[00:36:32]Then C is 10 km on a bearing of 140 from B.

[00:36:37]D is 10 km on a bearing of 160 from A.

[00:36:42]And in part A we are told using a scale of 1 cm to represent 2 km, show the relative positions of A, B, C, and D.

[00:36:53]Show the relative positions. Meaning that we are supposed to first locate the first reference point. B is on a bearing of 045 from A. Meaning that in our location A should come first. From A. A very key statement.

[00:37:12]So, allow me draw a compass, then I say uh that is point A.

[00:37:17]With point A a bearing of 045 means this is a true bearing. True bearing is supposed to be measured strictly from the north and in the clockwise direction.

[00:37:30]So, from the north and in the clockwise direction uh-huh, from the north.

[00:37:38]From the north.

[00:37:41]Sorry.

[00:37:42]From the north and in the clockwise direction.

[00:37:47]So, we identify 45 at this point.

[00:37:50]Then we produce that line.

[00:37:58]Uh-huh.

[00:37:59]This is exactly 45.

[00:38:04]Then can we check at this point now? Can we check the distance again?

[00:38:09]1 cm for 2 km. Meaning that 12 shall be represented by 6 because each centimeter is representing 2 km. Therefore, we measure 6.

[00:38:20]We measure 6.

[00:38:22]So, from point A 6 cm, then we make an arc.

[00:38:34]6 cm for the 12 km, we make an arc. Then after identifying point B, we raise a compass at that point, a very straight one, and parallel to the first one.

[00:38:46]At this point, we can now proceed to check the location of another place, C, 10 km on a bearing of 140 from B.

[00:38:57]A bearing of 140 from B. So, at B again, from the north, and in the clockwise direction, and in the clockwise direction, we identify 140, and we produce a straight line through that angle of 140.

[00:39:17]Through that angle of 140.

[00:39:19]Then now, after identifying such an angle of 140, along the same same angle, because it was 10 km, 10 shall be represented by five centimeters.

[00:39:32]Yeah, by 5 cm.

[00:39:35]Uh-huh. So, this is 5 cm to represent the 10 km. We raise again a compass at that particular point, a compass that is parallel to the first one, like that.

[00:39:48]Very upright. This becomes our point C.

[00:39:52]That becomes our point C.

[00:39:54]Uh-huh. We've measured 140, an angle of 140 like that. Again, we are told that a point D is 10 km on a bearing of 160 from A. So, from A, bearing of 160.

[00:40:12]Bearing of 160°.

[00:40:14]That is a true bearing. So, from the north and in the clockwise, we measure 160.

[00:40:21]Then we put a mark there and we produce a line.

[00:40:28]Uh-huh.

[00:40:30]We've measured 160, meaning that we only remain with the 20 here.

[00:40:35]I'm measuring 160.

[00:40:37]So now, we consider the distance again.

[00:40:39]It is a 10 km. So, we can talk about five now, 5 cm.

[00:40:47]So, you measure 5 cm along that.

[00:40:53]And remember, you're supposed to raise a compass.

[00:40:56]Pick up points.

[00:41:00]You raise a compass there.

[00:41:02]This becomes our point B.

[00:41:04]So, we have a the relative positions.

[00:41:06]We've been told that these are boundaries. So, we can join the border DC.

[00:41:13]The border DC, we can join it.

[00:41:18]Uh-huh, like that.

[00:41:20]Now, maybe very perfect.

[00:41:23]I connect.

[00:41:30]Yeah, like that.

[00:41:31]So, we've connected DC.

[00:41:33]Then now, can we check part B of the same same question. We've been told in part B, from the scale drawing, determine the compass bearing of D from C. Underline from from C and underline that it is compass direction. When we say compass direction, it means it can be from any of the main points, either north or south, and it has to be it has to be an acute angle. Therefore, when you measure it from the south, in case you connect D through the anticlockwise, it shall not be an acute angle. But connecting C through here, it's an acute angle. So, we measure this acute angle and you realize that it is 48°. So, we are moving from south through 48° and you're moving to the west.

[00:42:32]So, that is how we state compass bearing.

[00:42:36]A very good question. Remember, we can be allowed an accuracy of 1 degree plus or minus 1. So, if the right answer is 48, then people with 49 and 47 are considered to have done the right thing.

[00:42:53]But I encourage you to be perfectly correct. You can try out the same diagram.

[00:42:58]Let us see what you get on your book.

[00:42:59]Read the instructions carefully. Try to interpret and come up with answers just as I have done. So, that's a very good uh question. Can we check number nine together, ladies and gentlemen?

[00:43:13]In number nine, the question says that uh the gradient of a line L that passes through points A 2M4 and B -14 -1M, sorry, is 1/7. Find the equation of the line perpendicular to L at point B. That means for us to proceed, we have to find the value of M.

[00:43:38]Gradient, which is 1/7, is given by change in Y, so M - 4 out of change in X -1 -2M.

[00:43:52]Yes.

[00:43:54]-1 - 2M.

[00:43:56]So, now, can we carry out cross multiplication, whereby 7 * M -4 then 1 * -1 - 2M.

[00:44:10]At this point we have 7m - 28 = -1 - 2m.

[00:44:17]Can we talk of 9m being equal to 27 because we have -1 and 28 becomes positive on the right hand side. So -1 + 28 it becomes 27. So the value of m, ladies and gentlemen, uh the value of m shall exactly be three.

[00:44:40]So I've divided by nine both sides. If m is three, can we talk about specific coordinates now?

[00:44:49]A shall be 2m 4. So that is six.

[00:44:55]The coordinates of B shall be -1 m, which is a three.

[00:45:01]So we have coordinates now at this point.

[00:45:05]Now with these coordinates, we can now express the equation of a line perpendicular to L at point B. So a line is passing here and it's perpendicular to L.

[00:45:18]Perpendicular means it has a different gradient. We know that gradient one times gradient two should always give us -1 if lines are perpendicular.

[00:45:29]So gradient one is 1/7 * unknown gradient two should give us -1.

[00:45:36]Can we talk about gradient two now becoming -7? Yeah.

[00:45:42]Gradient two becomes -7.

[00:45:46]If gradient two is -7, if gradient two is -7, then we can find its equation. We also know that it passes through point B.

[00:45:56]So -7 = to change in y y - 3 out of change in x x minus negative one which makes it a positive one. Now, we can perform again cross multiplication whereby y minus three multiplies one and negative seven multiplies x and also one such that when negative seven multiplies x, it becomes negative seven x. Negative seven by one, it remains negative seven.

[00:46:29]At this point, now we can remain with y equals to negative seven x.

[00:46:35]Then negative seven plus three it becomes negative four.

[00:46:42]So, this is now the equation.

[00:46:45]This is now the equation of the second line which is perpendicular to L.

[00:46:53]A very good question.

[00:46:55]A very, very good question.

[00:47:01]Number 10.

[00:47:03]These are very good questions. If you are very keen then you will bear me witness that you guys are learning something.

[00:47:11]A very excellent paper.

[00:47:13]Remember to join for more.

[00:47:15]Remember to subscribe for more classes because we continue preparing you until the final day of your KCSE and KCPE respectively.

[00:47:27]Now, at number 10 Yeah, we can read number 10 together. We try to evaluate it together.

[00:47:34]Then we see.

[00:47:36]Then we see.

[00:47:37]So, in number 10 we told that two straight paths are perpendicular to each other at point P.

[00:47:49]All right? You know the meaning of perpendicular. One path meets a straight road at point A while the other meets the same rod at point B.

[00:48:02]Given that PA equals to 50, while PB equals to 60 m, calculate the obtuse angle made by PB and the rod.

[00:48:17]This is a very good question, which you're supposed to first analyze Aha. And understand what it means by everything.

[00:48:29]An obtuse angle is an angle that is between 90 and 180. Greater than 90 and less than 180. That's an obtuse angle.

[00:48:40]Obtuse angle. So, two straight paths here.

[00:48:44]This one and this one, they are perpendicular. When we say perpendicular, if this is P, they meet at P, and they are perpendicular to each other at point P. Now, one path meets the straight road at point A, while the other the same road at point B. Given that from P to A, it's 50. So, this can be point A.

[00:49:16]Then, P to B, it's 60.

[00:49:19]There is a road here.

[00:49:22]There is a road at this point.

[00:49:25]There is a road at that point. That's a road. 60 to B.

[00:49:31]50, like that.

[00:49:33]So, what is the question?

[00:49:35]We find the obtuse angle made by PB and the road.

[00:49:45]PB and the road. So, when PB meets the road, when PB PB can proceed like when they meet the road we love an angle at this point.

[00:49:57]Shall have an angle at that particular point.

[00:50:02]Therefore, we can first apply trigonometry and find the size of the angle at this point.

[00:50:12]Let me call it angle alpha. So, we can say that Let's Let me use tan.

[00:50:19]tan of angle alpha is given by the opposite length out of the adjacent length.

[00:50:26]Then now the angle itself will be given by the tan inverse of 5 over 6 or 50 over 60, which is the same.

[00:50:34]So, the angle shall be given by shift tan.

[00:50:40]I'm saying tan inverse, sorry. The angle will be given by tan inverse. So, shift tan 5 out of 6.

[00:50:49]I'm getting exactly 39.81.

[00:50:54]39.81 degrees.

[00:51:00]39.81 degrees.

[00:51:03]That is an acute angle. What about the obtuse angle?

[00:51:07]What about the obtuse angle? Because PB can be produced to the other side.

[00:51:11]So, what about this angle now? The obtuse one.

[00:51:15]So, theta will be given by They are on a straight line. So, 180 minus 39.81.

[00:51:24]This is going to be 180 minus I'm getting 140.19.19 degrees.

[00:51:36]So, that is the value of the angle obtuse for that case that we wanted at this point. It's a very good question.

[00:51:45]A very good question.

[00:51:47]Mhm. Actually, I like this paper.

[00:51:50]Personally, I like this paper because you may not know when it is it is a simple trigonometric ratios that are being assessed.

[00:52:01]A very technical paper whereby you're supposed to read and sketch information. Mhm? You read information and you sketch it until you come up with the right with the right answers. So, a very good paper.

[00:52:18]Don't forget to share it with your friends.

[00:52:21]Huh? How will they be friends if you can come across a paper like this one and you forget to just share with them. 11.

[00:52:29]A two-digit number is such that the sum of the ones and the twice the sum of the ones that is ones digit and twice the tens digit twice the tens digit is 16.

[00:52:44]When the digits are reversed, the number formed exceeds the original number by 36.

[00:52:51]So, the first thing is to form two simplified simultaneous equations to represent the above information. The two-digit number can be XY. These are the digits. So, when we take the ones digit which is Y and we add twice the tens digit we supposed to get 16.

[00:53:17]Question one.

[00:53:18]We together in that?

[00:53:21]Y plus, you know, Y is the one on the ones.

[00:53:24]Twice the tens digit.

[00:53:26]So, 2Y uh 2X should get 16.

[00:53:30]Then again, when the digits are reversed, there's something called total value of a number. The total value of a number is the number times the place it occupies. For example, original number has a total value of 10 * x y * 1.

[00:53:46]Now, the new number which is y x now shall be it is a y on the 10's position, then x on the ones. Now, this number formed exceeds the original by 36. So, when we pick the new number and we subtract the original number we are supposed to get 36 exactly.

[00:54:12]So, this is 10y + x - 10x - y = 36. Can we simplify the equation? Yes.

[00:54:23]10y and -y, we get 9y.

[00:54:27]x - 10x, get -9x.

[00:54:32]This is equal to 36. This is a good equation, but it can be simplified to y - x = Is it four?

[00:54:42]This is now the second equation, simplified like that. So, with the two simplified equations now, we can proceed we can proceed to part B of the question. Using matrix method solve the equations in A above and hence find the number.

[00:55:03]So, we have the first equation y + 2x = 16. The second one y - x = 4.

[00:55:15]So, we can extract the matrix equation from the coefficients and say this is one two one negative one. We call this one the coefficients matrix.

[00:55:31]Then times y x This is equal to 4.

[00:55:39]So, we are now having what you're calling the matrix equation made of the coefficients matrix, variables matrix, and the constants matrix.

[00:55:50]The second step is to ensure that we find the inverse.

[00:55:53]We find the inverse.

[00:55:55]Then that is a the inverse of the coefficients matrix.

[00:56:01]1 2 1 -1 So, let me I'm looking for a space.

[00:56:06]1 2 1 -1 So, we can first find the dates.

[00:56:12]Determinant shall be 1 by -1 because it should be product of elements in the leading diagonal minus product of elements in the other diagonal.

[00:56:23]So, this is -1 -2, which is -3. So, the determinant is -3. Then now for us to get the inverse, we're supposed to exchange positions for elements in the leading diagonal such that -1 comes here and 1 on the other.

[00:56:39]Then two changes the sign, and also one changes the sign. So, elements in the other diagonal, this diagonal, will just change their sign, but for the leading diagonal, they interchange positions.

[00:56:54]Then now we multiply or we divide everything.

[00:56:58]We now divide everything by by the determinant.

[00:57:04]We divide everything by the determinant.

[00:57:08]Yeah, like that.

[00:57:09]So, when we divide by the determinant, it means multiplying by its reciprocal.

[00:57:14]To divide by -3, multiply by It means multiplying by -1/3. So, -1/3 * -1, it becomes 1/3. -1/3 * -2, it becomes positive 2/3. -1/3 * -1, it becomes 1/3.

[00:57:30]Then -1/3 * 1, it becomes negative 1/3.

[00:57:39]Yeah, so we are having the inverse of the matrix. With the inverse of the matrix now, we can pre-multiply our matrix equation by this particular inverse.

[00:57:47]A third, negative a third, two thirds.

[00:57:52]So we now having a third, two thirds, a third again. I think it's a third there.

[00:58:03]Then negative a third.

[00:58:06]Like that. Let me check whether I've transferred the right thing.

[00:58:11]A thirds are here. Then two thirds and negative a third.

[00:58:16]A thirds, two thirds, negative a third, like that.

[00:58:19]Uh-huh. So we are pre-multiplying both sides. So one, two, one, negative one.

[00:58:27]Ahem, Y X according to the order, the variables, equals to we start with a third, two thirds, a third, negative a third, multiplied by 16, four, like that. The other thing that you supposed to understand, ladies and gentlemen, is that a matrix times its own inverse, it's a matrix and its own inverse, we get the identity matrix. Identity matrix is a matrix that has one in the main diagonal and zeros in the other diagonal.

[00:59:08]So that is automatic. You don't even have to perform the multiplication. Then now on the other side, we perform multiplication. Row times column. So we start with a third, 16, uh-huh, plus two dots multiplied by four.

[00:59:30]I'm getting exactly eight.

[00:59:32]That point.

[00:59:34]Then the next row by the same same column.

[00:59:39]I'll call you.

[00:59:41]Mhm, we start with a third.

[00:59:45]Into 16.

[00:59:48]The negative a third into four.

[00:59:53]This is giving me exactly four.

[00:59:57]The other thing that you supposed to know is that any matrix times the identity matrix remains unchanged.

[01:00:05]So, X Okay, YX. Sorry.

[01:00:10]So, Y X remains YX because YX has been multiplied by the identity matrix. This is equal to eight.

[01:00:19]Four. Then from this now, we can get the values of Y as eight, the values of X as four, and therefore to find the number.

[01:00:36]Huh?

[01:00:37]Now, we write the number is whereby the original number was XY, then it becomes 48 now. So, the number is 48.

[01:00:47]48 is the number.

[01:00:50]48 is the number. That's a very good question. 48 is the number.

[01:00:56]Can we check now number 12? Yes, number 12.

[01:01:00]Uh-huh.

[01:01:01]In number 12, we are told that the figure below shows a pyramid with a square base of side six and vertical height 14. So, the vertical height is 14.

[01:01:14]Yeah.

[01:01:16]The base is a square. Find the total surface area.

[01:01:20]Total surface area.

[01:01:22]we can find first area of all the faces, then we add.

[01:01:29]From V to Y From V to Y, can we find V to Y? Remember this are the middle at the center. So, from O to Y, it is a three.

[01:01:44]Because up to the end here it is six. So, that is a three. 14 and three, you see Pythagoras theorem will give us the length V to Y. So, we need 14 squared plus three squared.

[01:02:00]Then now we take the square roots.

[01:02:02]Mhm.

[01:02:04]So, I need 14 squared plus three squared.

[01:02:09]We take square roots. This is 14.32.

[01:02:16]And these are centimeters.

[01:02:19]14.32.

[01:02:21]Therefore, with 14.32 being a square based pyramid, you can know that uh You can know very well that the slanting faces are all equal.

[01:02:34]Therefore, they are triangular. Half times a base of six and a height of 14.32.

[01:02:42]Then after that, we shall multiply by four such faces. So, half times six times the answer here.

[01:02:51]I'm getting 42.95.

[01:02:57]Then we multiply by four because they are four such heights.

[01:03:03]And this will give us a 42.95 by four, I'm getting 171.8.

[01:03:17]Square centimeters. That is not from the sloping triangular faces.

[01:03:22]The now the base shall be six by six giving us 36 square centimeters.

[01:03:32]Then the total now total surface area shall be the base plus the other slanting faces.

[01:03:42]Uh-huh, when we add and we can two added in seven point eight square centimeters.

[01:03:53]So, we have now the total surface area. The total surface area has been obtained. This is a very good question.

[01:04:06]Very good question.

[01:04:08]Very good question.

[01:04:11]Can we now check number 13?

[01:04:14]In number 13, we told that uh the figure below shows a velocity time graph for a particle in motion for 48 seconds.

[01:04:25]Find the maximum speed attained by the particle in kilometers per hour if the total distance covered by the particle is 728 meters. So, there's a speed attained at this point.

[01:04:42]There's a speed there.

[01:04:44]This speed X attained.

[01:04:47]From the bottom here, the value of X can be the height there.

[01:04:52]So, one thing that we supposed to know is that one thing that we must understand even before you do anything else is that uh area covered equals to the Okay, distance covered equals to the area under the graph of velocity time.

[01:05:12]Therefore, what do we use to get the area here?

[01:05:16]To get the area, we need to Okay, it has already been partitioned.

[01:05:21]Though we can still continue partitioning.

[01:05:26]Though we can still continue partitioning, we can use area one, area two, area three. So, area one is a trapezoidal trapezium in nature.

[01:05:36]Mhm.

[01:05:37]Being trapezium, how would we get area?

[01:05:40]Area one, half multiplied by sum of the two parallel sides. From zero all the way to X, then from zero again all the way to eight.

[01:05:52]Then the height between them from zero to 12.

[01:05:56]Uh-huh. Can we simplify this? Yes.

[01:06:00]So, I'm getting 6 X and 48.

[01:06:05]What about area two?

[01:06:07]Rectangular.

[01:06:08]From zero all the way to X up there, multiplied by from 12 all the way to 32.

[01:06:15]From 12 to 32, that is a division of 20. So, 20 by X.

[01:06:20]This is now going to be area two.

[01:06:22]Then area three is triangular. So, half times a base of from that two to 48.

[01:06:31]This is 16 and a height of X.

[01:06:35]So, this shall be 8 X. So, the total area gives us the distance covered.

[01:06:43]So, 6 X plus 48 plus 20 X plus 8 X, we should get 728.

[01:06:56]Can we add up?

[01:06:58]26 plus Let's 6 plus 20. Then we add eight.

[01:07:04]34 X.

[01:07:07]This is equivalent to 728.

[01:07:11]We subtract 48 and we get exactly 680.

[01:07:19]Yeah, 680.

[01:07:21]So, to remain with x then, we need to divide by 34 both sides and this is exactly 20 m/s.

[01:07:30]So, the velocity, the highest velocity attained is 20 m/s. Now, 20 m/s ought to be converted to km/h according to the instructions.

[01:07:43]km/h. So, leaving it at that point shall not attract you the final mark.

[01:07:48]So, we convert now.

[01:07:51]20 m should be divided by 1,000 to become kilometers divided by seconds. 1 second should be divided by 3600 to become hours.

[01:08:10]So, this is a 20 over 1,000 multiplied by 3600 over 1.

[01:08:18]This shall be 20 by 3600 out of 1,000.

[01:08:25]I'm getting exactly 72 km/h.

[01:08:32]I know you have very many techniques of converting velocity or speed from km/h to m/s and vice versa.

[01:08:43]So, you can use your own method to come up with the same answer.

[01:08:49]Number 14.

[01:08:50]Yeah, can we use number 14 now?

[01:08:53]As we wind up. Number 14, a very good question. In number 14, we're being told one thing.

[01:09:00]Being told one thing that we find the scalars m and n such that m multiplied by 4 3 then n multiplied by negative 3 2 equals to 5 8. So you see the x component we can talk about a 4 m then n by -3 that becomes -3n equals to 5. This is equation 1 and also on the y components this is a 3 m then n by 2 that is a 2 n equals to 8. So you see a second equation. Then now we solve the simultaneous equations. I know there are many methods of solving simultaneous equations. Allow me solve using substitution from equation 2 2n equals to 8 minus 3m then the value of n becomes 4 minus 1.5 m.

[01:10:03]So we replace this now in equation 1.

[01:10:06]Equation 1 becomes 4m minus 3 and instead of n 4 minus 1.5 m this is equal to 5.

[01:10:17]So 4m minus 12 then positive 4.5 m equals to 5. Grouping terms together shall give us a 8 0.5 m equals to 17. So the value of m shall be what?

[01:10:38]17 divide by 8.5 giving me exactly 2. So the scalar multiple m the scalar m is a 2 and n can be found by from these are 4 minus 1.5 into 2. This is 4 - 3 in other words.

[01:11:02]So we get exactly one.

[01:11:05]4 - 3 get one.

[01:11:08]So the value of n is one.

[01:11:10]The value of m is two and we are done with a scalar multiples. A very good question.

[01:11:20]15 Juma needs to import a car from Japan.

[01:11:24]Whose cost is $5,000.

[01:11:28]Now he intends to buy the car through an agent who deals in Japanese Yen.

[01:11:34]Right?

[01:11:35]The agent will charge him 20% commission of the price of the car and further 80,325 Japanese Yen for shipment of the car.

[01:11:47]All right. Calculate the amount to the nearest Kenya shillings he'll be spending to obtain the car given the following exchange rates.

[01:11:58]This is a very good question whereby we supposed to know first.

[01:12:04]You can see the cost of the car is in a US dollars. All right.

[01:12:10]And if it's in the US dollars then at the end of the day we need all the costs all the costs we need them to be in Japanese Yen.

[01:12:24]This particular guy called Juma is using an agent whereby the agent needs 20% of the price of the car. Therefore, commission commission shall be 20% of 5,000 US dollars.

[01:12:47]20% of 5,000 gives us exactly 1,000.

[01:12:58]Yeah?

[01:12:59]20% equals to 1,000.

[01:13:03]So, with this now we can see that now at the end of the day it will cost Juma 5,000.

[01:13:14]And again 1,000 will be paid to the agent.

[01:13:20]These are US uh US dollars. It's a US dollars.

[01:13:25]This is what it will cost Juma.

[01:13:27]Again shipment exercise.

[01:13:30]For shipment uh shipment here.

[01:13:36]Shipment.

[01:13:38]Therefore we ask ourselves, how can we convert this Japanese yen also how can we convert uh the Japanese yen also into the US dollars or the US dollars to Japanese yen. So, let's convert this US dollars to Japanese yen by first making it uh by first making it what?

[01:14:07]Uh by first making it Kenya shillings.

[01:14:10]So, we say that uh converting foreign currency to Kenyan currency.

[01:14:19]The bank buys yeah, the bank shall buy.

[01:14:22]Therefore, we say that uh to Kenyan shillings the amount will be bought at 127.

[01:14:33]Therefore 6,000 by 127 this gives us 7 62.

[01:14:43]Kenyan shillings.

[01:14:44]762 Kenyan shillings.

[01:14:46]Therefore, uh this particular guy this particular guy called Juma for the cost of the car plus for the cost of the car then plus the commission he needs to give to the agent should be totaling to 762.

[01:15:15]762,000.

[01:15:18]Yeah.

[01:15:19]Then apart from that apart from that we are also supposed to ask ourselves.

[01:15:27]Supposed to ask ourselves that You know there is a Japanese yen here which should also be converted to shillings. So we convert also these ones to shillings so that you may add and know what Juma should be having.

[01:15:41]What Juma should be having. Therefore, we will say this, eh?

[01:15:47]If 100 Japanese yen I equal to this amount respectively, what about just one Japanese yen?

[01:15:54]You divide by 100 across.

[01:15:57]8427 0.91 53. I've divided by 100 to get one Japanese yen.

[01:16:05]Therefore, we want to convert Japanese yen to Kenya shillings. That is foreign to Kenya.

[01:16:13]Foreign to Kenya we again use the column and on the buying column it will be 80 325 times the amount at which Japanese yen is bought.

[01:16:30]84 27 8427 So, 80 325 by 0.

[01:16:40]8427 So, I'm getting Kenya Shillings 67.

[01:16:47]I have 689.

[01:16:51]689 point 88.

[01:16:58]And remember we want it in nearest Kenya Shillings. So, 67 690 Kenya Shillings. This is now for shipment. Meaning that the total that he should be having the total that he should be having in Kenya Shillings should be should be Aha.

[01:17:24]Just a minute.

[01:17:26]Just a minute. I want us to flow together at this point.

[01:17:31]I want us to flow together at this point.

[01:17:35]Just to be patient with me, Ndung'u Ti.

[01:17:38]I'm still on the interpretation.

[01:17:41]You know we are saying this.

[01:17:43]You should be having some Kenyan Shilling.

[01:17:46]And you know how to convert Kenyan Shillings to foreign we use the selling.

[01:17:50]Yeah, Kenya Kforce. We use the selling column by division.

[01:17:55]So, you should be having some Kenyan Shillings such that when they are divided by 132 it becomes 6,000 US Dollars.

[01:18:06]Yes.

[01:18:07]There's some Kenyan Shillings that we don't know such that when converted to US it becomes 6,000.

[01:18:13]I think that is the concept you should use at that point. So, X Kenya Shillings should be 132 by 6,000.

[01:18:23]So, we get 792,000.

[01:18:27]This is the amount in Kenya Shillings that he should have be having for the commission and also for the car such that when we convert it now to US dollars, we get 6,000. Then again, we should be having some amount in Kenya shillings at this point.

[01:18:46]We should be having some amount in Kenya shillings such that if converted if converted to Japanese, it will become 80,325.

[01:18:56]So again, unknown amount should be divided by 0.91 53 and give us 8,000 325. Meaning that Y the shipment amount now should be 80 325 multiplied by 0.91 53.

[01:19:21]So this is giving us 73,000 521 point 47.

[01:19:31]Meaning that now the total amount that he should be having in Kenya shillings should be when we add now 792 thousand.

[01:19:45]So this is going to be 86 I know 865 521. And remember we've been told to the nearest shillings. So this now becomes the final answer at that point.

[01:20:00]It's a very good question, but should be interpreted with a lot of care and keenness.

[01:20:06]Then number 16, which is a question on a Number 16 is a question on the usage of tables.

[01:20:15]Yeah, the usage of tables.

[01:20:17]So we can check it out.

[01:20:20]We can check it out and work it correctly. Work it correctly.

[01:20:28]Uh-huh.

[01:20:33]Use tables of reciprocals.

[01:20:36]Use tables of reciprocals and square roots to evaluate and square roots to evaluate three out of Mhm.

[01:20:46]Three out of 0.5 to 1 plus the root of 0.43 4036.

[01:20:55]A very good uh question. So, we can check it now.

[01:21:00]We can check it.

[01:21:01]I want us to open the tables and also organize our work very well uh in this particular question.

[01:21:11]This particular question.

[01:21:16]Mhm.

[01:21:19]So, we're supposed to know that uh from the reciprocals, we're having three multiplied by the reciprocal now.

[01:21:28]Five to one.

[01:21:30]Then we add the square root of You know, for us to find square root, we've to write in standard form and ensure that the power of 10 is the power of 10 is even. So, we are finding the square root of This becomes instead of 4.0, we make it 40.36 * 10 ^ 2 10 ^ -2.

[01:21:54]So, now we can use reciprocals first, but before we go to the reciprocals table, allow me write 5.21 * 10 to the power of -1 uh huh in standard form, then plus the square root. The square root of the first part should be read from the tables. So, we are reading the square root of 40.36 first. 40.36.

[01:22:22]So, we check the four-figure tables.

[01:22:24]40.36.

[01:22:29]Mhm.

[01:22:31]Square roots, yeah?

[01:22:33]40.36 So, 40 is here.

[01:22:38]40 is here. We are on the tables of square roots, right?

[01:22:41]So, 40.36 So, 40.3 Then, six. Six should be here. 40.3 then six.

[01:22:50]So, 40.3 on that column, this is what you find.

[01:22:56]So, 6.

[01:22:58]uh 3482 6.3482 3482 Uh-huh. That is a 40.3 because it will be 36. Now, we go to the addition part where there is six.

[01:23:14]We check along that column of six.

[01:23:17]We find 47 here. So, we are adding 47 at this point.

[01:23:21]Uh-huh.

[01:23:23]This is going to be nine.

[01:23:25]It's going to be 12, so carry one.

[01:23:27]Five, three.

[01:23:29]Six.

[01:23:31]3529 6.3529 6.3 five 29 times When we are finding square roots now, the power divides by two.

[01:23:47]When we are finding square roots, the power divides by two.

[01:23:50]Then, now here, three times We also check the reciprocal of 5.21.

[01:23:58]Reciprocal of 5.21 So, we check the reciprocal tables.

[01:24:03]The reciprocal tables Uh-huh.

[01:24:09]Reciprocal of 5.

[01:24:11]5.21 So, 5.2 Then, one is here. So, 5.21 shall be 0.

[01:24:21]We start it 0 here.

[01:24:23]1919. 0.1919 1919 * 10 power The reciprocal of this part becomes 10 with the power but with the the reversed sign. If it is -1, the reciprocal becomes +1. Then now this part you can say the square root is 0. We multiply by -1, 10 ^ -1 63 5 29 Then here we multiply by 3.

[01:24:57]Then also * 10 ^ 1.

[01:24:59]So we multiply and get exactly 5.7 57 0.635 5 29 So when we add this is becoming exactly 6.

[01:25:24]39 2 29 So this is now the final answer.

[01:25:31]In number 16 from tables of reciprocals and square roots, what you're supposed to know is that with reciprocals the power of 10 changes the sign.

[01:25:44]So it became +1.

[01:25:46]Then with square root the power of 10 divides by 2. And that's why in your expression to standard form, you ensure that the power is even so that you don't get any challenges when dividing by 2.

[01:25:59]So it became -1 from -2. Then now you compile the question and get your answer. This is a very good question.

[01:26:09]Later we'll be looking at the section two of the same paper. Thank you so much for following.

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