WAEC SSCE - General Mathematics - 2022

Akọwa Nkọwa

To calculate the gradient of the line, we need to find the slope, which is represented by "m" in the equation y = mx - 4. The slope of a line tells us how steep or flat it is. To find the slope, we can use the coordinates of the given point (-4, 16) and the equation of the line. The equation tells us that for any point on the line, the y-coordinate (vertical) is equal to the slope multiplied by the x-coordinate (horizontal) minus 4. Let's substitute the given point's coordinates into the equation: 16 = m(-4) - 4 Now, let's simplify the equation: 16 = -4m - 4 To solve for "m," we need to isolate it on one side of the equation. Let's add 4 to both sides: 16 + 4 = -4m Simplifying further: 20 = -4m To find the value of "m," we divide both sides by -4: 20/-4 = m Simplifying the division: -5 = m Therefore, the gradient or slope of the line is -5.

Akọwa Nkọwa

Convert from base 5 to base 10

432five ${}_{\mathrm{<br>}}$ =  (4 x 52 ${}^2$) + (3 x 51 ${}^1$) + (2 x 50 ${}^0$)

= (4 x 25) + (3 x 5) + (2 x 1)

= 100 + 15 + 2

= 117ten ${}_{\mathrm{<br>}}$

Then convert from base 10 to base 3

Selecting the remainders from bottom to top:

117ten ${}_{\mathrm{<br>}}$ = 11100three ${}_{\mathrm{<br>}}$

Hence; 432five ${}_{\mathrm{<br>}}$ =  11100three

Akọwa Nkọwa

1x $\frac{\mathrm{<br>}}{}$ +43x $\frac{\mathrm{<br>}}{}$ - 56x $\frac{\mathrm{<br>}}{}$ + 1 = 0
using 6x as lcm

→ 6+8−5+6x6x $\frac{\mathrm{<br>}}{}$ 

→ 9+6x6x $\frac{\mathrm{<br>}}{}$ = 0

9+6x = 0

6x = -9

x = −96 $\frac{<br>}{}$  or −32

Akọwa Nkọwa

We can use the property that an angle inscribed in a circle is half the measure of the arc it intercepts. Since triangle MNR is inscribed in circle MNR, we know that angle MNR is half the measure of arc MR. Similarly, angle MRN is half the measure of arc MN. We also know that arc PQM and arc QNR add up to a full circle, which has a measure of 360 degrees. Using these facts, we can write two equations: - angle MNR = 1/2(arc MR) = 1/2(360 - arc PQM - arc QNR) - angle MRN = 1/2(arc MN) = 1/2(360 - arc PQM) Substituting the given angle measures, we have: - 41 = 180 - arc PQM/2 - arc QNR/2 - 141 = 180 - arc PQM/2 Solving for arc PQM in the second equation gives us arc PQM = 198. Substituting this into the first equation, we have: 41 = 180 - 99 - arc QNR/2 arc QNR/2 = 40 arc QNR = 80 Therefore, angle QNR is half the measure of arc QNR, which is 80 degrees. So, the answer is 80º.

Akọwa Nkọwa

43x ${}^{\mathrm{<br>}}$ = 42(x+1) ${}^{\mathrm{<br>}}$

3x = 2x + 2

3x - 2x = 2

x = 2

Akọwa Nkọwa

Let's assume the cost price of the article to be 100. As per the question, the trader made a loss of 15% when the article was sold. Therefore, the selling price would be: Selling price = Cost price - Loss Selling price = 100 - 15% of 100 Selling price = 100 - 15 Selling price = 85 So, the ratio of selling price to cost price can be calculated as: Selling price : Cost price = 85 : 100 Selling price : Cost price = 17 : 20 Therefore, the ratio of the selling price to cost price is 17:20. Hence, the correct option is (C) 17:20.

Akọwa Nkọwa

using the trial method by inserting each option in the equation.

Inserting 21º: sin([5 x 21] - 28) = cos([3 x 21] - 50)

sin(105 - 28) = cos (63 - 50)

sin 77º = cos 13º 

where:

sin 77º = 0.9744

cos 13º =  0.9744

Akọwa Nkọwa

To find the probability that Mrs. Gabriel will give birth to a girl with blue eyes, we can multiply the probabilities of each event. Given: Probability of giving birth to a girl = 1/2 Probability of having a baby with blue eyes = 1/4 To find the probability of both events happening, we multiply the individual probabilities: Probability of girl with blue eyes = (Probability of girl) * (Probability of blue eyes) = (1/2) * (1/4) = 1/8 Therefore, the probability that Mrs. Gabriel will give birth to a girl with blue eyes is 1/8. So, the correct answer is 1/8.

Akọwa Nkọwa

Let's start by using the formula for the perimeter of a rectangle: P = 2L + 2W, where L is the length and W is the width. We are given that the length L is 10 cm and the perimeter P is 28 cm. So we can substitute these values into the formula and solve for the width W: P = 2L + 2W 28 = 2(10) + 2W 28 = 20 + 2W 2W = 8 W = 4 Now that we know the width of the rectangle is 4 cm, we can use the formula for the area of a rectangle: A = L x W, where L is the length and W is the width. We already know L is 10 cm and we just found that W is 4 cm. So we can substitute these values into the formula and solve for the area A: A = L x W A = 10 x 4 A = 40 Therefore, the area of the rectangle is 40 cm2.

Akọwa Nkọwa

573.06 x 184.25 = 105,586.305

1,05600.00 to four significant figure

What are the Rules for significant figures?

Significant Figures

• All non-zero numbers ARE significant.
• Zeros between two non-zero digits ARE significant. 
• Leading zeros are NOT significant.
• Trailing zeros to the right of the decimal ARE significant.
• Trailing zeros in a whole number with the decimal shown ARE significant.

Akọwa Nkọwa

The weaver bought a bundle of grass for $50.00 and made 8 mats from it. Therefore, the cost of each mat is $50.00/8 = $6.25. If each mat is sold for $15.00, then the profit per mat is $15.00 - $6.25 = $8.75. To find the percentage profit, we need to calculate the profit as a percentage of the cost. Profit percentage = (Profit/Cost) x 100% Profit percentage = ($8.75/$6.25) x 100% Profit percentage = 140% Therefore, the weaver made a 140% profit on the sale of each mat. The answer is (b) 140%.

Akọwa Nkọwa

To solve the equation 6x^2 = 5x - 1, we first move all the terms to one side to obtain a quadratic equation in standard form, which is 6x^2 - 5x + 1 = 0. Then we can apply the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a, where a = 6, b = -5, and c = 1. Plugging these values into the quadratic formula, we get: x = (-(-5) ± sqrt((-5)^2 - 4(6)(1))) / 2(6) x = (5 ± sqrt(25 - 24)) / 12 x = (5 ± 1) / 12 Therefore, the solutions to the equation are x = 1/2 and x = 1/3. So the answer is x = 1/2, 1/3.

Akọwa Nkọwa

2−18m21+3m $\frac{\mathrm{<br>}}{}$ =  2[1−9m2]1+3m $\frac{\mathrm{<br>}}{}$

2[1−3m][1+3m]1+3m $\frac{\mathrm{<br>}}{}$

2[1-3m]

Akọwa Nkọwa

The formula for the volume of a cone is V = 1/3 πr^2h, where r is the radius of the base and h is the height of the cone. Substituting the given values into the formula: V = 1/3 × 22/7 × (3.5cm)^2 × 12cm = 1/3 × 22/7 × 12.25cm^2 × 12cm = 1/3 × 22/7 × 147cm^3 = 22/7 × 49cm^3 = 154cm^3 Therefore, the volume of the cone is 154cm^3. The answer is (d) 154cm3.

Akọwa Nkọwa

The mean of two numbers x and y is 4, which means (x + y) / 2 = 4. Solving for x + y, we get x + y = 8. Now we need to find the mean of four numbers x, 2x, y, and 2y. The mean is calculated by adding up all the numbers and dividing by the total number of numbers. So the sum of these four numbers is x + 2x + y + 2y = 3x + 3y, and the total number of numbers is 4. Therefore, the mean is (3x + 3y) / 4. We can substitute 8 for x + y since we know it equals 8, and simplify the expression to get (3x + 3y) / 4 = (3(x+y)) / 4 = (3(8)) / 4 = 6. Therefore, the mean of the four numbers x, 2x, y, and 2y is 6. The answer is option C.

Akọwa Nkọwa

For a regular polygon with n sides, each exterior angle is equal to 360°/n. Given that the exterior angle of the regular polygon is 22.5°, we can set up an equation as follows: 360°/n = 22.5° Multiplying both sides by n, we get: 360° = 22.5°n Dividing both sides by 22.5°, we get: n = 360°/22.5° = 16 Therefore, the regular polygon has 16 sides. Hence, the answer is (d) 16.

Akọwa Nkọwa

The sum of angles in a triangle = 180º

(180 - 108)º = 72º

Isosceles triangle has both two equal sides and two equal angles.

 ∠OSN =  ∠SON = 722 $\frac{\mathrm{<br>}}{}$

= 36º

Akọwa Nkọwa

To find the largest share, we need to first add the ratio values to get the total parts in the sharing. The sum of the ratio values is 6+7+8 = 21. This means that the sharing is divided into 21 parts. To find the largest share, we need to determine the fraction of the sharing that each boy receives. The first boy receives 6 out of 21 parts, which is (6/21) of the total sharing. The second boy receives 7 out of 21 parts, which is (7/21) of the total sharing. The third boy receives 8 out of 21 parts, which is (8/21) of the total sharing. To find the amount of money that each boy receives, we divide the total amount by 21 and multiply by the respective ratio value. The largest share will be received by the boy with the largest ratio value, which is the third boy who receives 8 out of 21 parts. So, the largest share is: 8/21 * 10,500 = 4,000 Therefore, the correct answer is (a) 4000.

Akọwa Nkọwa

θ360
  * 2 * π * r → 150∗2∗22∗4.2360x7 $\frac{\mathrm{<br>}}{}$

= 11cm

Akọwa Nkọwa

The problem tells us that we have a set of 10 numbers, and that the average (or mean) of all those numbers is 56. However, we also know that if we exclude the 10th number from the set and find the average of the remaining 9 numbers, the result is 55. To find the value of the 10th number, we need to use a little bit of algebra. Let's call the sum of the first nine numbers "S". We know that the average of those nine numbers is 55, so: S / 9 = 55 If we multiply both sides of the equation by 9, we get: S = 495 Now we can use the fact that the average of all ten numbers is 56 to find the 10th number. The sum of all ten numbers is: 10 x 56 = 560 We already know that the sum of the first nine numbers is 495, so we can subtract that from the total to get the value of the 10th number: 560 - 495 = 65 Therefore, the 10th number in the set is 65.

Akọwa Nkọwa

In ΔWYZ: 

hyp2 ${}^2$ = adj2 ${}^{\mathrm{<br>}}$ + opp2 ${}^{\mathrm{<br>}}$

hyp2 ${}^{\mathrm{<br>}}$ =32 ${}^{\mathrm{<br>}}$ + 42 ${}^{\mathrm{<br>}}$ → 9 + 16

hyp2 ${}^{\mathrm{<br>}}$ = 25

hyp = 5

In  ΔWXY:

hyp2 ${}^{\mathrm{<br>}}$ = adj2 ${}^{\mathrm{<br>}}$ + opp2 ${}^{\mathrm{<br>}}$

hyp2 ${}^{\mathrm{<br>}}$ = 122 ${}^{\mathrm{<br>}}$ + 52 ${}^{\mathrm{<br>}}$ = 144 +25

hyp2 ${}^{\mathrm{<br>}}$ = 169

hyp = 13

the perimeter of WXYZ. = 3+4+12+13 → 32cm

Akọwa Nkọwa

The surface area of an open-top cylinder = πr(r + 2h),

where 'r' is the radius and 'h' is the height of the cylinder.

= 227 $\frac{\mathrm{<br>}}{}$ * 3.5 (3.5 + 2 * 8)

= 11 (3.5 + 16) → 11 (19.5)

= 214.5cm2

Akọwa Nkọwa

Error = 1.80 - 1.75 = 0.05

%error = errororiginallength $\frac{\mathrm{<br>}}{}$

=  0.05∗1001.75 $\frac{\mathrm{<br>}}{}$

= 2.85 or  267

Akọwa Nkọwa

 5b+(a+b)2(a−b)2 $\frac{\mathrm{<br>}}{}$

= 5∗−7+(3+−7)2(3−−7)2 $\frac{\mathrm{<br>}}{}$

=  −35+16102 $\frac{<br>}{}$

=  −19100 $\frac{<br>}{}$ or -0.19

Akọwa Nkọwa

Let's use algebra to solve the problem. Given that Mary has $3.00 more than Ben, we can express Ben's amount of money in terms of x as (x - 3). Also, given that Mary has $5.00 less than Jane, we can express Jane's amount of money in terms of x as (x + 5). Therefore, the sum of Ben and Jane's money would be: Ben + Jane = (x - 3) + (x + 5) Simplifying this expression, we get: Ben + Jane = 2x + 2 So, the correct answer is $(2x+2).

Akọwa Nkọwa

The angle of elevation

= Tan θ = oppadj $\frac{\mathrm{<br>}}{}$

Tan θ = 12+1410 $\frac{\mathrm{<br>}}{}$

Tan θ =  2610 $\frac{\mathrm{<br>}}{}$

θ = Tan−1 ${}^{<br>}$(2.6)

θ ≈ 70º

Akọwa Nkọwa

To evaluate 23 x 54 (mod 7), we need to perform the multiplication first, which gives us 1242. Then, we take this number modulo 7 by finding the remainder when 1242 is divided by 7. Dividing 1242 by 7 gives us a quotient of 177 and a remainder of 3. Therefore, 1242 is congruent to 3 (mod 7). So, 23 x 54 (mod 7) is equivalent to 3 (mod 7). Therefore, the answer is 3.

Akọwa Nkọwa

n(AuB) = n(A) + n(B) -  n(AnB)

n(AuB) = 8 + 12 - 3

= 17 

Akọwa Nkọwa

To find the lower quartile, we need to first arrange the ages in ascending order. Here are the ages provided: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41, and 36. Next, we divide the data set into four equal parts, with each part containing an equal number of values. The lower quartile is the median of the first half of the data set. Let's arrange the ages in ascending order: 12, 13, 15, 36, 39, 41, 41, 43, 43, 47, 49 We can see that there are 11 ages in total, which means the first half contains 11/2 = 5.5 ages. Since we can't have a fraction of an age, we round down to the nearest whole number, which is 5. Now, let's look at the five ages in the first half of the data set: 12, 13, 15, 36, 39. To find the lower quartile, we need to find the median of this subset of ages. Since there are an odd number of ages (5), the median is the middle value. In this case, the middle value is 15. Therefore, the lower quartile is 15. So, the correct answer is 15.

Akọwa Nkọwa

To find the perpendicular distance between the parallel sides of a trapezium, we can use the formula for the area of a trapezium. The formula for the area of a trapezium is given by: Area = (1/2) * (sum of parallel sides) * (perpendicular distance between the parallel sides) In this case, we are given: Length of one parallel side = 9 cm Length of the other parallel side = 12 cm Area of the trapezium = 105 cm² Let's denote the perpendicular distance between the parallel sides as "h." Using the formula for the area of a trapezium, we can rewrite it as: 105 = (1/2) * (9 + 12) * h Simplifying the equation: 105 = (1/2) * 21 * h Multiplying both sides by 2 to eliminate the fraction: 210 = 21h Dividing both sides by 21 to solve for "h": h = 210 / 21 h = 10 cm Therefore, the perpendicular distance between the parallel sides of the trapezium is 10 cm. So, the correct answer is 10 cm.

Akọwa Nkọwa

The sum of angles in a triangle = 180º

(180 - 108)º = 72º

Isosceles triangle has both two equal sides and two equal angles.

 ∠OSN =  ∠SON = 722 $\frac{\mathrm{<br>}}{}$

= 36º

Akọwa Nkọwa

The question presents two quadratic equations: x^2 - 5x + 6 = 0 and x + px + 6 = 0. Both equations are said to have the same roots. To find the value of p, we need to determine when two quadratic equations have the same roots. For two quadratic equations, ax^2 + bx + c = 0 and dx^2 + ex + f = 0, to have the same roots, the following conditions must be met: 1. The discriminant of both equations must be zero. 2. The ratios of the corresponding coefficients must be equal. Let's apply these conditions to the given equations: Equation 1: x^2 - 5x + 6 = 0 Equation 2: x + px + 6 = 0 1. Discriminant condition: The discriminant of Equation 1 is given by Δ1 = b1^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1. The discriminant of Equation 2 is given by Δ2 = e^2 - 4df = p^2 - 4(1)(6) = p^2 - 24. Since both equations have the same roots, their discriminants must be equal: Δ1 = Δ2 1 = p^2 - 24 p^2 = 25 p = ±5 2. Ratio of coefficients condition: Comparing the coefficients of the corresponding terms: For Equation 1: a1 = 1, b1 = -5, c1 = 6 For Equation 2: a2 = 1, b2 = p, c2 = 6 The ratio of the coefficients must be equal: a1/a2 = b1/b2 = c1/c2 From a1/a2 = 1/1 = 1 and c1/c2 = 6/6 = 1, we can conclude that: b1/b2 = -5/p = 1 -5/p = 1 p = -5 Therefore, the value of p that satisfies both conditions and makes the equations have the same roots is p = -5.

Akọwa Nkọwa

The problem tells us that M varies directly as n and inversely as the square of p. This can be written as: M ∝ n/p^2 Where the symbol "∝" means "is proportional to". We can also write this using a constant of proportionality k: M = k(n/p^2) To find the value of k, we can use the values of M, n, and p given in the problem: 3 = k(2/1^2) Simplifying this equation, we get: k = 3/2 Now we can use this value of k to find M in terms of n and p: M = (3/2)(n/p^2) Simplifying further: M = (3n)/(2p^2) Therefore, the answer is: - 3n/(2p^2)

Akọwa Nkọwa

The area of a rectangle = length ✕ width

: Length = area ÷ width → 3 38 $\frac{\mathrm{<br>}}{}$ ÷ 34 $\frac{\mathrm{<br>}}{}$

= 278 $\frac{\mathrm{<br>}}{}$ * 43 $\frac{\mathrm{<br>}}{}$

= 412 $\frac{\mathrm{<br>}}{}$cm

Akọwa Nkọwa

Recall that: log3 ${}_3$ 27 → log3 ${}_3$33 ${}^{\mathrm{<br>}}$

3log3 ${}_3$3 → 3 * 1

= 3

Then log3 ${}_3$ 27 = 2x + 1

→ 3 = 2x + 1 

3 - 1 = 2x

2 = 2x

1 = x

Akọwa Nkọwa

To find the mean of a set of numbers, you need to add up all the numbers and then divide by the total number of numbers. In this case, we have 11 numbers: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41, 36 To find the mean, we add up all these numbers: 12 + 47 + 49 + 15 + 43 + 41 + 13 + 39 + 43 + 41 + 36 = 379 Then we divide by the total number of numbers, which is 11: 379 ÷ 11 = 34.45 Therefore, the mean of the set is 34.45.

Akọwa Nkọwa

multiply both sides by the LCM

12 x y+24 $\frac{\mathrm{<br>}}{}$ -  12 x y−13 $\frac{\mathrm{<br>}}{}$ >12 x 1

3[y+2] - 4[y-1] > 12

3y + 6 - 4y + 4 > 12

-y + 2 > 12

-y > 12 - 10 = 2

y < -2

Akọwa Nkọwa

The formula for the volume of a sphere is V = (4/3)πr^3, where r is the radius of the sphere and π is the constant pi (approximately equal to 3.14). Given that the radius of the sphere is 3 cm, we can substitute the value of r into the formula to get: V = (4/3) × 22/7 × 3^3 = (4/3) × 22/7 × 27 ≈ 113.14 cm^3 Therefore, the correct option is 113.14cm3, rounded to two decimal places.

Akọwa Nkọwa

To find the value of y in the given expression, let's simplify it step by step: √24 + √96 - √600 = y√6 First, let's simplify the square roots inside the expression: √24 = √(4 * 6) = √4 * √6 = 2√6 √96 = √(16 * 6) = √16 * √6 = 4√6 √600 = √(100 * 6) = √100 * √6 = 10√6 Now, we substitute the simplified values back into the expression: 2√6 + 4√6 - 10√6 = y√6 Combining like terms: (2 + 4 - 10)√6 = y√6 Simplifying further: -4√6 = y√6 Since the coefficient of √6 on both sides of the equation is the same, we can conclude that the value of y must be -4. Therefore, y = -4. So, the correct answer is -4.

Akọwa Nkọwa

To find the probability that a household reads at least one paper, we need to add the probabilities of households that read the Morning Times and those that read the Evening Dispatch, and then subtract the probability of households that read both papers since they are being counted twice. Let's use the following notation: - M = the event that a household reads the Morning Times - E = the event that a household reads the Evening Dispatch We are given that: - P(M) = 0.45 (45% of households read the Morning Times) - P(E) = 0.60 (60% of households read the Evening Dispatch) - P(M ∩ E) = 0.20 (20% of households read both papers) To find the probability that a household reads at least one paper, we can use the formula: P(M ∪ E) = P(M) + P(E) - P(M ∩ E) Substituting the values we have: P(M ∪ E) = 0.45 + 0.60 - 0.20 P(M ∪ E) = 0.85 Therefore, the probability that a particular household reads at least one paper is 0.85 or 85%.

Akọwa Nkọwa

The given statements are: - p: Stephen is intelligent - q: Stephen is good at Mathematics The symbol "⇒" means "implies". So, p⇒q means "if Stephen is intelligent, then he is good at Mathematics". , "If Stephen is good at Mathematics, then he is intelligent", is a valid conclusion because it is the contrapositive of p⇒q. In other words, if the original statement is true, then its contrapositive must also be true. The contrapositive of p⇒q is "if Stephen is not good at Mathematics, then he is not intelligent" However, is also valid because the contrapositive of a true statement is always true. , "If Stephen is not intelligent, then he is not good at Mathematics", is the inverse of the original statement and is not necessarily true. In other words, just because Stephen is not intelligent, it does not mean that he is not good at Mathematics. For example, he may have a natural talent for Mathematics even if he is not generally intelligent. , "If Stephen is not good at Mathematics, then he is intelligent", is the converse of the original statement and is also not necessarily true. In other words, just because Stephen is good at Mathematics, it does not mean that he is intelligent. For example, he may be good at Mathematics but struggle with other subjects. So, the valid conclusion is: "If Stephen is good at Mathematics, then he is intelligent".

Akọwa Nkọwa

θ360
  *π * r2 ${}^{\mathrm{<br>}}$ → 210∗22∗4.2∗4.2360∗7 $\frac{\mathrm{<br>}}{}$

161750 $\frac{\mathrm{<br>}}{}$ = 32.34cm2

Akọwa Nkọwa

5x + 3y=4

5x-3y= 2

Using elimination method 

5x + 3y=4 →    5x + 3y=4

-[5x-3y= 2] → -5x +3y= -2

6y = 2 

y → 1/3 and x = 3/5

solving (25x2 ${}^{\mathrm{<br>}}$ -9y2 ${}^{\mathrm{<br>}}$)

25 * [3/5]2 ${}^{\mathrm{<br>}}$ -9 * [1/3]2 ${}^{\mathrm{<br>}}$

25 * 925 $\frac{\mathrm{<br>}}{}$ - 9 19

9 - 1 = 8

Akọwa Nkọwa

square both sides to remove the square root

k2 ${}^2$ = m2 ${}^2$ t−pr $\frac{\mathrm{<br>}}{}$

k2rm2 $\frac{{\mathrm{<br>}}^{}}{}$ = t - p

t = k2rm2 $\frac{{\mathrm{<br>}}^{}}{}$ + p

t = k2r+pm2m2 $\frac{{\mathrm{<br>}}^{}}{}$ 

Akọwa Nkọwa

(x2 ${}^{\mathrm{<br>}}$ - 4x + p)

Use the coefficient of the middle variable(-4x)

= (−42
)2 ${}^{\mathrm{<br>}}$

= (-2)2 ${}^{\mathrm{<br>}}$

= 4

Akọwa Nkọwa

To find the 17th term of the arithmetic progression -6, -1, 4, we need to first find the common difference, d, between the terms: d = -1 - (-6) = 5 Now we can use the formula for the nth term of an arithmetic progression: a_n = a_1 + (n - 1) * d where a_1 is the first term and n is the term we want to find. Plugging in the values we have: a_17 = -6 + (17 - 1) * 5 a_17 = -6 + 16 * 5 a_17 = -6 + 80 a_17 = 74 Therefore, the 17th term of the arithmetic progression is 74. Answer option (C) is correct.

Akọwa Nkọwa

To find how high up the wall the ladder reaches, we can use trigonometry, specifically the sine function. Given: Length of the ladder = 6m Angle between the ladder and the horizontal = 53º We want to find the height of the ladder on the wall. Using the trigonometric relationship for right triangles, we can use the sine function to relate the angle and the sides of the triangle. sin(angle) = opposite / hypotenuse In this case, the height of the ladder on the wall is the opposite side, and the length of the ladder is the hypotenuse. sin(53º) = height / 6 To find the height, we rearrange the equation: height = sin(53º) * 6 Using a calculator, we can evaluate sin(53º) ≈ 0.7986. height ≈ 0.7986 * 6 ≈ 4.7916 Therefore, the height up the wall that the ladder reaches is approximately 4.7916m. So, the correct answer is 4.792m.

Akọwa Nkọwa

(√3−3√3 $\frac{<br>}{}$) = √3√3 $\frac{<br>}{}$\

3−√33 $\frac{\mathrm{<br>}}{}$

= 1 - √3

20 = 15 + 10 - x

x = 25 - 20 = 5

P(both) = 520 $\frac{\mathrm{<br>}}{}$

= 14 $\frac{\mathrm{<br>}}{}$ or 0.25

Akọwa Nkọwa

(a) ∠XYZ = 12 $\frac{1}{2}$ (146) = 73º

∠OZX = 34º

∠OZY = 73 - 34

∠OYZ = ∠OZY = 39  

(b) x = y - 7

42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360

122°+7x = 360° 7x 238°

x = 34

: 34 = y - 7

34 + 7 = y

41 = y