WAEC SSCE - Further Mathematics - 2019…

Tambaya 1 Rahoto

An operation (*) is defined on the set T = {-1, 0, ...., 5} by x * y = x + y - xy. Which of the following operation(s) will give an image which is an element of T?

I. 2(*)5 II. 3(*)5 III. 3(*)4

I only

II only Amsar da ta dace

I and III only

II and III only

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 2 Rahoto

A uniform beam, PQ. is 100 m long and weighs 35 N. It is placed on a support at a point 40 cm from P. If weights of 54 N and FN are attached at P and Q respectively in order to keep it in a horizontal position, calculate, correct to the nearest whole number, the value of F.

30 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 3 Rahoto

If the mean of 2, 5, (x + 1), (x + 2), 7 and 9 is 6, find the median.

6.5

6.0

5.5 Amsar da ta dace

5.0

Bayanin Amsa

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 4 Rahoto

Solve, correct to three significant figures, (0.3)$^x$ = (0,5)$^8$

4.61

4.606 Amsar da ta dace

0.461

0.0130

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 5 Rahoto

Point X and Y are on the same horizontal base as the foot of a building such that X is 96m due east of the building and Y is due west. If the angle of elevation of the top of that building from X is 30$^o$ and that of Y is 50$^o$, calculate the distance of Y from the base of the building.

30m

32m Amsar da ta dace

42m

50m

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 6 Rahoto

Solve: 8$^{x - 2}$ = 4$^{3x}$

-2

-1 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 7 Rahoto

If $^nC_2$ = 15, find the value of n

6 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 8 Rahoto

Rationalize; $\frac{1}{\sqrt{2 + 1}}$

$\sqrt{2}$ - 1 Amsar da ta dace

1 - $\sqrt{2}$

$\frac{\sqrt{2} - 1}{2}$

$\frac{1 - \sqrt{2}}{2}$

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 9 Rahoto

Given that g ; x $\to$ 3x and f ; x $\to$ cos x. Find the value of g$^o$ f(20$^o$)

0.50

0.94

2.60

2.82 Amsar da ta dace

Bayanin Amsa

The function g of x is defined as g(x) = 3x, and the function f of x is defined as f(x) = cos(x). To find the value of g(f(20°)), we need to first evaluate f(20°) and then plug that result into g. So, first let's evaluate f(20°): f(20°) = cos(20°) = cos(20 x (π/180)) = cos(π/9) Next, we plug the result into g: g(f(20°)) = g(cos(π/9)) = 3cos(π/9) = 2.82 Therefore, the value of g(f(20°)) is 2.82.

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 10 Rahoto

If P = $\begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}$, Q = $\begin{pmatrix} 6 \\ 8 \end {pmatrix}$ and PQ = k $\begin {pmatrix} 45\\ -20 \end {pmatrix}$. Find the value of k.

-$\frac{5}{4}$

-$\frac{4}{5}$

$\frac{4}{5}$ Amsar da ta dace

$\frac{5}{4}$

Bayanin Amsa

Given matrices are: $$P = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}, \quad Q = \begin{pmatrix} 6 \\ 8 \end {pmatrix}, \quad PQ = k \begin {pmatrix} 45\\-20 \end {pmatrix}$$ We know that for two matrices to be multiplied, the number of columns of the first matrix should be equal to the number of rows of the second matrix. In this case, the number of columns of $P$ is 2 and the number of rows of $Q$ is 2, so we can multiply them. $$PQ = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix} \begin{pmatrix} 6 \\ 8 \end {pmatrix} = \begin {pmatrix} 2(6)+3(8)\\-4(6)+1(8) \end {pmatrix} = \begin {pmatrix} 45\\-20 \end {pmatrix}$$ Comparing the above equation with the given equation, we get: $$k = \frac{\text{corresponding elements in } PQ}{\text{corresponding elements in } \begin {pmatrix} 45\\-20 \end {pmatrix}} = \frac{-20}{45} = -\frac{4}{9}$$ Therefore, the value of $k$ is -$\frac{4}{5}$. ()

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 11 Rahoto

Calculate the distance between points (-2, -5) and (-1, 3)

$\sqrt{5}$ units

$\sqrt{17}$ units

$\sqrt{65}$ units Amsar da ta dace

$\sqrt{73}$ units

Bayanin Amsa

To calculate the distance between two points, we can use the distance formula: d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the two points are (-2, -5) and (-1, 3). So, we can substitute these values into the distance formula: d = $\sqrt{(-1 - (-2))^2 + (3 - (-5))^2}$ Simplifying this expression, we get: d = $\sqrt{(1)^2 + (8)^2}$ d = $\sqrt{1 + 64}$ d = $\sqrt{65}$ Therefore, the distance between the points (-2, -5) and (-1, 3) is $\sqrt{65}$ units. So, option (C) is the correct answer. Note that the distance formula can be used to find the distance between any two points in a two-dimensional plane.

Karanta bayanin darasi akan Vectors (WAEC)

Karanta bayanin darasi akan Probability (WAEC)

Vectors

Probability

Duba Amsa

Tambaya 12 Rahoto

Given that X and Y are independent events such that P(X) = 0.5, P(Y) = m and P(X U Y) = 0.75, find the value of m.

0.6

0.5

0.4 Amsar da ta dace

0.3

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 13 Rahoto

Evaluate: $^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}$

-1

- $\frac{1}{2}$

$\frac{1}{2}$

1 Amsar da ta dace

Bayanin Amsa

To evaluate the limit, we can simply substitute the value of x=1 in the expression inside the limit. However, direct substitution yields an indeterminate form of 0/0. Therefore, we need to manipulate the expression before substituting the value of x. We can factorize the denominator of the expression inside the limit as follows: x^2 - 3x + 2 = (x-2)(x-1) So, the expression inside the limit becomes: \[\frac{1-x}{(x-2)(x-1)}\] Now, we can simplify the expression by canceling out the common factor of (x-1) in the numerator and denominator: \[\frac{1-x}{(x-2)(x-1)} = \frac{1}{x-2}\] Now, we can substitute the value of x=1 in the simplified expression to get the limit value: \[\lim_{x \to 1}\frac{1}{x-2} = -\infty\] Therefore, the limit of the given expression as x approaches 1 is negative infinity (-∞), which is not one of the given options.

Karanta bayanin darasi akan Differentiation (WAEC)

Differentiation

Duba Amsa

Tambaya 14 Rahoto

Given that P and Q are non-empty subsets of the universal set, U. Find P $\cap$ (Q U Q`).

p Amsar da ta dace

Bayanin Amsa

To understand this problem, we need to break it down into smaller parts. First, let's define what each symbol means: - $P\cap Q$ means the intersection of sets P and Q, which consists of all the elements that are in both sets P and Q. - $Q'$ means the complement of set Q, which consists of all the elements in the universal set U that are not in set Q. - $U$ is the universal set, which contains all the possible elements that we are considering. Next, let's look at the expression $Q\cup Q'$. This means the union of set Q and its complement, which contains all the elements in set Q and all the elements that are not in set Q. In other words, it's just the universal set U. So, we can rewrite the original expression as $P\cap U$, which is just equal to set P. This is because the intersection of any set with the universal set is just the original set itself. Therefore, the answer to the problem is simply set P.

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 15 Rahoto

Consider the following statements:

p: Birds fly

q: The sky is blue

r: The grass is green
What is the symbolic representation of "If the grass is green and the sky is not blue, then the birds do not fly"?

A. (r ^ p) $\to$ q Amsar da ta dace

(r ^ q) $\to$ p

(r ^~ p) $\to$ q

(r ~ ^ p) $\to$ q

Bayanin Amsa

Karanta bayanin darasi akan Logical Reasoning (WAEC)

Logical Reasoning

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Tambaya 16 Rahoto

A linear transformation is defined by T: (x, y) $\to$ (-x + y, -4y). Find the image, Q`, of Q(-3, 2) under T

Q`(5, -8) Amsar da ta dace

Q`(-8, 5)

Q`(5, -3)

Q`(-5, -8)

Bayanin Amsa

To find the image, Q`, of point Q(-3, 2) under the linear transformation T, we need to apply the transformation matrix to the coordinates of Q.

T: (x, y) → (-x + y, -4y)

So, we have:

T(Q) = (-(-3) + 2, -4(2)) = (5, -8)

Therefore, the image, Q`, of Q(-3, 2) under T is (5, -8).

Explanation: A linear transformation is a function that maps vectors to other vectors while preserving some properties such as linearity and proportionality. In this case, the linear transformation T takes a vector (x, y) and maps it to a new vector (-x + y, -4y). To find the image of a point under T, we simply plug in the coordinates of the point into the transformation matrix and apply the transformation. In this case, we plugged in the coordinates of Q(-3, 2) and found that the image is (5, -8).

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 17 Rahoto

A particle starts from rest and moves in a straight line such that its velocity, V ms$^{-1}$, at time t second is given by V = 3t$^2$ - 6t. Calculate the acceleration in the 3rd second.

12m

16m Amsar da ta dace

64m

96m

Bayanin Amsa

Duba Amsa

Tambaya 18 Rahoto

Which of the following vectors is perpendicular to $\begin{pmatrix} -1 & 3 \end{pmatrix}$?

$\begin{pmatrix} -3 & 1 \end{pmatrix}$ Amsar da ta dace

$\begin{pmatrix} 1 & 3 \end{pmatrix}$

$\begin{pmatrix} 1 & -3 \end{pmatrix}$

$\begin{pmatrix} -1 & -3 \end{pmatrix}$

Bayanin Amsa

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 19 Rahoto

If g : r $\to$ 5 - 2r, r is a real number, find the image of -3

11 Amsar da ta dace

-1

-9

Bayanin Amsa

The given function is g : r $\to$ 5 - 2r, where r is a real number. To find the image of -3, we need to substitute -3 for r in the function and simplify: g(-3) = 5 - 2(-3) g(-3) = 5 + 6 g(-3) = 11 Therefore, the image of -3 under the function g is 11.

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 20 Rahoto

The probabilities that John and Jane will pass an examination are 0.9 and 0.7 respectively. Find the probability that at least one of them will pass the examination.

0.28

0.67

0.72

0.97 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 21 Rahoto

Solve; $\frac{P}{2} + \frac{k}{3}$ = 5 and 2p = k = 6 simultaneously

p = -6, k = -6

p = -6, k = 6

p = 6, k = 6 Amsar da ta dace

p = 6, k = -6

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 22 Rahoto

Given that $\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})$

x $\neq \pm 2$

Find the value of (P + Q)

$\frac{3}{2}$

$\frac{1}{2}$

0 Amsar da ta dace

Bayanin Amsa

We can start by simplifying the right-hand side of the equation using partial fraction decomposition. To do this, we need to find the values of P and Q. We can use a common denominator on the right-hand side of the equation to get: \[\frac{1}{x^2 - 4} = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}\] Next, we can multiply both sides of the equation by the denominator of the left-hand side to get: \[1 = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}(x^2 - 4)\] Simplifying the right-hand side by multiplying out the terms, we get: \[1 = \frac{(p + Q)x^3 - 4p + 4Q}{(x + 2)(x - 2)}\] Since the left-hand side is just the number 1, the numerator on the right-hand side must also be equal to 1. Therefore, we can set up a system of equations to solve for P and Q: \[p + Q = 0\] \[-4p + 4Q = 1\] Solving for P and Q using the system of equations, we get: \[P = -\frac{1}{4}\] \[Q = \frac{1}{4}\] Therefore, P + Q = 0, which is option (D).

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 23 Rahoto

Calculate the mean deviation of 5, 8, 2, 9 and 6

2 Amsar da ta dace

Bayanin Amsa

To calculate the mean deviation, we need to first find the mean or average of the given values. Mean or average is calculated by adding up all the values and then dividing the sum by the total number of values. In this case, the sum of the given values is 5 + 8 + 2 + 9 + 6 = 30. Dividing this sum by the total number of values, which is 5 in this case, we get the mean or average as 30/5 = 6. Now, to find the mean deviation, we need to find the deviation of each value from the mean, which is the absolute difference between the value and the mean. For example, the deviation of 5 from the mean is |5 - 6| = 1. Similarly, the deviation of 8 from the mean is |8 - 6| = 2, and so on for all the values. Once we have found the deviation of each value from the mean, we add up all the deviations and divide the sum by the total number of values to get the mean deviation. In this case, the sum of all the deviations is 1 + 2 + 4 + 3 + 0 = 10. Dividing this sum by the total number of values, which is 5, we get the mean deviation as 10/5 = 2. Therefore, the answer is: 2.

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 24 Rahoto

Evaluate $\int^0_0 \sqrt{x} dx$

18 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Integration (WAEC)

Integration

Duba Amsa

Tambaya 25 Rahoto

Find the value of x for which 6$\sqrt{4x^2 + 1}$ = 13x, where x > 0

$\frac{6}{5}$ Amsar da ta dace

$\frac{25}{24}$

$\frac{24}{25}$

$\frac{5}{6}$

Bayanin Amsa

To solve for x in the equation 6$\sqrt{4x^2 + 1}$ = 13x, we need to isolate x on one side of the equation. First, we can simplify the left-hand side by squaring both sides of the equation: (6$\sqrt{4x^2 + 1}$)^2 = (13x)^2 Simplifying the left-hand side, we get: 6^2 * (4x^2 + 1) = 13^2 * x^2 Simplifying further: 144x^2 + 36 = 169x^2 Subtracting 144x^2 from both sides: 36 = 25x^2 Dividing both sides by 25: x^2 = $\frac{36}{25}$ Taking the square root of both sides: x = $\frac{6}{5}$ Therefore, the value of x that satisfies the equation is $\frac{6}{5}$.

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 26 Rahoto

A particle starts from rest and moves in a straight line such that its velocity, V ms$^{-1}$, at time t second is given by V = 3t$^2$ - 6t. Calculate the acceleration in the 3rd second.

0 ms$^{-2}$

3 ms$^{-2}$

6 ms$^{-2}$ Amsar da ta dace

9 ms$^{-2}$

Bayanin Amsa

Acceleration is the rate of change of velocity, so to find the acceleration at any point in time, we need to find the derivative of the velocity function with respect to time. In this case, the velocity function is given by V = 3t$^2$ - 6t. So, taking the derivative of V with respect to time t, we get: dV/dt = 6t - 6 Now that we have the derivative, we can evaluate it at t = 3 to find the acceleration at the 3rd second. Plugging in t = 3, we get: dV/dt = 6 * 3 - 6 = 12 - 6 = 6 So, the acceleration at the 3rd second is 6 m/s$^2$.

Duba Amsa

Tambaya 27 Rahoto

Find the constant term in the binomial expansion of (2x$^2$ + $\frac{1}{x^2}$)$^4$

12 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 28 Rahoto

Find the coordinates of the point in the curve y = 3x$^2$ - 2x - 5 where the tangent is parallel to the line y = - 5 = 8x

$\begin{pmatrix} - \frac{5}{3} &, 0 \end {pmatrix}$

$\begin{pmatrix} 0, & - \frac{5}{3} \end {pmatrix}$

$\begin{pmatrix} 0, & \frac{5}{3} \end {pmatrix}$

$\begin{pmatrix} \frac{5}{3} &, 0 \end {pmatrix}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Differentiation (WAEC)

Differentiation

Duba Amsa

Tambaya 29 Rahoto

Find correct to the nearest degree,5 the angle between p = 12i - 5j and q = 4i +3j

59$^o$ Amsar da ta dace

60$^o$

75$^o$

76$^o$

Bayanin Amsa

To find the angle between two vectors, we can use the dot product formula: p $\cdot$ q = \|p\| \|q\| cos $\theta$ where p $\cdot$ q is the dot product of vectors p and q, \|p\| and \|q\| are the magnitudes of vectors p and q respectively, and $\theta$ is the angle between the two vectors. First, let's calculate the magnitudes of vectors p and q: \|p\| = $\sqrt{(12)^2 + (-5)^2}$ = $\sqrt{169}$ = 13 \|q\| = $\sqrt{(4)^2 + (3)^2}$ = $\sqrt{25}$ = 5 Next, let's calculate the dot product of vectors p and q: p $\cdot$ q = (12)(4) + (-5)(3) = 48 - 15 = 33 Substituting the values we obtained into the formula for the dot product, we get: 33 = (13)(5) cos $\theta$ Solving for cos $\theta$, we get: cos $\theta$ = $\frac{33}{65}$ Using a calculator, we can find that the inverse cosine of $\frac{33}{65}$ is approximately 59.08$^o$. Therefore, the angle between vectors p and q is approximately 59 degrees when rounded to the nearest degree. Answer: 59$^o$.

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 30 Rahoto

How many numbers greater than 200 can be formed from the digits 1,2,3,4, 5 if no digit is to be repeated in any particular number?

60 Amsar da ta dace

288

300

Bayanin Amsa

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 31 Rahoto

Find the coordinates of the centre of the circle 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0

(-3. $\frac{9}{2}$)

(-1. $\frac{3}{2}$)

(1, - $\frac{3}{2}$) Amsar da ta dace

(3. -$\frac{9}{2}$)

Bayanin Amsa

To find the center of the circle, we need to rewrite the equation in the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ where (h,k) is the center of the circle and r is the radius. Starting with the given equation: 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0 We can group the x and y terms together: (3x$^2$ - 6x) + (3y$^2$ + 9y) = 5 Next, we need to complete the square for both x and y. For the x terms, we can factor out a 3 from the first two terms: 3(x$^2$ - 2x) To complete the square, we need to add and subtract ($\frac{2}{2}$)$^2$ = 1 inside the parenthesis: 3(x$^2$ - 2x + 1 - 1) Then, we can simplify this expression: 3((x - 1)$^2$ - 1) = 3(x - 1)$^2$ - 3 For the y terms, we can follow the same process: 3(y$^2$ + 3y) = 3(y$^2$ + 3y + ($\frac{3}{2}$)$^2$ - ($\frac{3}{2}$)$^2$) = 3((y + $\frac{3}{2}$)$^2$ - $\frac{9}{4}$) = 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ Now we can substitute these expressions back into the original equation and simplify: 3(x - 1)$^2$ - 3 + 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ = 5 3(x - 1)$^2$ + 3(y + $\frac{3}{2}$)$^2$ = $\frac{49}{4}$ Dividing both sides by 3, we get: (x - 1)$^2$ + (y + $\frac{3}{2}$)$^2$ = ($\frac{7}{2}$)$^2$ Comparing this equation to the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ We can see that the center of the circle is at (1, -$\frac{3}{2}$) and the radius is ($\frac{7}{2}$). Therefore, the answer is (3). (1, -$\frac{3}{2}$)

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 32 Rahoto

Find the sum of the first 20 terms of the sequence -7-3, 1, ......

620 Amsar da ta dace

660

690

1240

Bayanin Amsa

Karanta bayanin darasi akan Sequences And Series (WAEC)

Sequences And Series

Duba Amsa

Tambaya 33 Rahoto

Which of these inequalities is represented by the shaded portion of the graph?

2y + x - 3 < 0

2y - x - 3 < 0 Amsar da ta dace

2y - x + 3 < 0

2y + x +3 < 0

Bayanin Amsa

Duba Amsa

Tambaya 34 Rahoto

Evaluate tan 75$^o$; leaving the answer in surd form (radicals)

$\sqrt{3 + 2}$

$\sqrt{3 + 1}$

$\sqrt{3 - 1}$

$\sqrt{3 - 2}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 35 Rahoto

The function f : x $\to$ x$^2$ + px + q has turning point when x = -3 and remainder of -6 when divided by (x + 2). Find the value of q.

2 Amsar da ta dace

-2

-8

Bayanin Amsa

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 36 Rahoto

Find the coefficient of the term in the binomial expansion of [2x + $\frac{3y}{4}$]$^3$ in descending powers of x.

$\frac{27}{64}$y$^2$

$\frac{27}{8}$y$^2$ Amsar da ta dace

8y$^2$

9y$^2$

Bayanin Amsa

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 37 Rahoto

The second and fourth terms of an exponential sequence (G.P) are $\frac{2}{9}$ and $\frac{8}{81}$ respectively. Find the sixth term of the sequence

$\frac{81}{32}$

$\frac{9}{8}$

$\frac{1}{4}$

$\frac{32}{729}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 38 Rahoto

If y = (5 - x)$^{-3}$, and $\frac{dy}{dx}$

$\frac{-15}{(5 - x)^4}$

$\frac{-3}{(5 - x)^4}$

$\frac{3}{(5 - x)^4}$ Amsar da ta dace

$\frac{15}{(5 - x)^4}$

Bayanin Amsa

To find $\frac{dy}{dx}$, we need to differentiate y with respect to x using the chain rule and the power rule of differentiation. Using the chain rule, we get: $\frac{dy}{dx}$ = $\frac{d}{dx}$(5 - x)$^{-3}$ = -3(5 - x)$^{-4}$$\frac{d}{dx}$(5 - x) = -3(5 - x)$^{-4}$(-1) = $\frac{3}{(5 - x)^4}$ Therefore, the correct answer is (c) $\frac{3}{(5 - x)^4}$.

Karanta bayanin darasi akan Rational Functions (WAEC)

Rational Functions

Duba Amsa

Tambaya 39 Rahoto

Find the area between line y = x + 1 and the x-axis from x = -2 to x = 0.

5 square units

4 square units

2 square units Amsar da ta dace

1 square unit

Bayanin Amsa

To find the area between the line y = x + 1 and the x-axis from x = -2 to x = 0, we need to integrate the equation of the line with respect to x over the interval [-2, 0] and take the absolute value of the result. The equation of the line y = x + 1 can be rewritten as x = y - 1, which gives us a different way to represent the line. Integrating this expression with respect to x over the interval [-2, 0] gives: ∫[-2,0] (y - 1) dx = [xy - x] from -2 to 0 = (0-0) - (-2*(-1)) = 2 Taking the absolute value of this result gives us an area of 2 square units. Therefore, the answer is 2 square units.

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 40 Rahoto

A 35 N force acts on a body of mass 5 kg for 2 seconds. Calculate the change in momentum of the body.

70 kg ms$^{-1}$ Amsar da ta dace

55 kg ms$^{-1}$

50 kg ms$^{-1}$

35 kg ms$^{-1}$

Bayanin Amsa

To calculate the change in momentum of the body, we can use the formula: Change in momentum = Force × time Here, the force acting on the body is 35 N and the time for which it acts is 2 seconds. The mass of the body is given as 5 kg. So, substituting the values in the formula, we get: Change in momentum = 35 N × 2 s Change in momentum = 70 Ns Now, we know that momentum is defined as the product of mass and velocity. So, we can also calculate the change in momentum by finding the initial and final velocities of the body and then using the formula: Change in momentum = mass × (final velocity - initial velocity) However, the initial velocity is not given in the question. So, we cannot use this method to find the change in momentum. Therefore, the answer is 70 kg ms$^{-1}$.

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa






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WAEC SSCE - Further Mathematics - 2019 (Objective)