WAEC SSCE - Further Mathematics - 2010…

Vraag 1 Verslag

Calculate, correct to one decimal place, the length of the line joining points X(3, 5) and Y(5, 1).

4.0

4.2

4.5 Juist antwoord

5.0

Antwoorddetails

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Vraag 2 Verslag

What is the angle between $a = (3i - 4j)$ and $b = (6i + 4j)$?

13°

87° Juist antwoord

100°

110°

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Trigonometry

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Vraag 3 Verslag

Find the maximum value of $2 + \sin (\theta + 25)$.

3 Juist antwoord

Antwoorddetails

The value of $\sin(\theta + 25)$ varies between -1 and 1. Therefore, the maximum value of $2 + \sin(\theta + 25)$ occurs when $\sin(\theta + 25) = 1$. This occurs when $\theta + 25 = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. Thus, $\theta = \frac{\pi}{2} - 25 + 2n\pi$. Substituting this value of $\theta$ into the expression for $2 + \sin(\theta + 25)$, we get: \[2 + \sin\left(\frac{\pi}{2} + 2n\pi\right) = 2 + 1 = 3\] Therefore, the maximum value of $2 + \sin(\theta + 25)$ is 3.

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Trigonometry

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Vraag 4 Verslag

Find the domain of $f(x) = \frac{x}{3 - x}, x \in R$, the set of real numbers.

${x : x \in R, x \neq 3}$ Juist antwoord

${x : x \in R, x \neq 1}$

${x : x \in R, x \neq 0}$

${x : x \in R, x\neq -3}$

Antwoorddetails

The function $f(x) = \frac{x}{3-x}$ is defined for all real numbers except when the denominator is zero, i.e., when $3-x=0$. Therefore, the domain of the function is all real numbers except $x=3$, or in interval notation: ${x : x \in R, x \neq 3}$. So the correct option is: ${x : x \in R, x \neq 3}$.

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Vraag 5 Verslag

Simplify $(1 + 2\sqrt{3})^{2} - (1 - 2\sqrt{3})^{2}$

$8\sqrt{3}$ Juist antwoord

$2 - 4\sqrt{3}$

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Surds

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Vraag 6 Verslag

Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in $ms^{-2}$, the magnitude of the acceleration of the body.

$\frac{\sqrt{2}}{5}$ Juist antwoord

$5\sqrt{2}$

$2\sqrt{5}$

$\frac{5\sqrt{2}}{2}$

Antwoorddetails

To find the magnitude of acceleration, we need to use Newton's Second Law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, we have two forces: (2i - 5j)N and (-3i + 4j)N. To find the net force, we add these two vectors by adding their corresponding components: (2i - 5j)N + (-3i + 4j)N = -i - j N So the net force is (-i - j)N. Now, using Newton's Second Law, we can find the acceleration: F = ma where F is the net force, m is the mass, and a is the acceleration. In our case, the net force is (-i - j)N and the mass is 5kg, so we have: (-i - j)N = 5kg * a Solving for a, we get: a = (-i - j)N / 5kg To find the magnitude of a, we take the square root of the sum of the squares of its components: |a| = sqrt((-1)^2 + (-1)^2) N/kg = sqrt(2) N/kg Converting to $ms^{-2}$ by dividing by 5, we get: |a| = sqrt(2)/5 $ms^{-2}$ Therefore, the answer is $\frac{\sqrt{2}}{5}$.

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Vraag 7 Verslag

The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.

12.4 years

12.6 years Juist antwoord

13.2 years

14.1 years

Antwoorddetails

Let's use the formula for the mean (average) of a set of numbers: \[\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}\] Let the age of the new pupil be x. Before the new pupil joined the class, the sum of ages of the 15 pupils would be: \[15 \times 14.2 = 213\] After the new pupil joined the class, the sum of ages of the 16 pupils would be: \[15 \times 14.2 + x = 213 + x\] We know that the new mean is 14.1, so we can set up an equation: \[\frac{15 \times 14.2 + x}{16} = 14.1\] Multiplying both sides by 16, we get: \[15 \times 14.2 + x = 16 \times 14.1\] Simplifying the left side, we get: \[213 + x = 225.6\] Subtracting 213 from both sides, we get: \[x = 12.6\] Therefore, the age of the new pupil is 12.6 years. Answer: (b) 12.6 years.

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Statistics

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Vraag 8 Verslag

For what values of x is $\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}$ undefined?

6 or 3

-18 or -9

-7 or 5 Juist antwoord

-5 or 7

Antwoorddetails

The expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we can find the values of x that make the denominator zero by setting it equal to zero and solving for x: $$x^{2} + 2x - 35 = 0$$ We can factor this quadratic equation as: $$(x + 7)(x - 5) = 0$$ So, the denominator is equal to zero when x = -7 or x = 5. Therefore, the expression is undefined for these values of x. So the correct answer is: - -7 or 5

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Rational Functions

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Vraag 9 Verslag

If $h(x) = x^{3} - \frac{1}{x^{3}}$, evaluate $h(a) - h(\frac{1}{a})$.

-1

$2a^{3} - \frac{2}{a^{3}}$ Juist antwoord

$\frac{2}{a^{3}} - 2a^{3}$

Antwoorddetails

We can start by computing the values of $h(a)$ and $h(\frac{1}{a})$, and then find their difference. First, we have $h(a) = a^{3} - \frac{1}{a^{3}}$. Next, we have $h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}$. Therefore, \begin{align*} h(a) - h(\frac{1}{a}) &= \left(a^{3} - \frac{1}{a^{3}}\right) - \left(\frac{1}{a^{3}} - a^{3}\right) \\ &= a^{3} - \frac{1}{a^{3}} - \frac{1}{a^{3}} + a^{3} \\ &= 2a^{3} - \frac{2}{a^{3}}. \end{align*} So the answer is $2a^{3} - \frac{2}{a^{3}}$. Therefore, is the correct answer.

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Polynomial Functions

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Vraag 10 Verslag

A bicycle wheel of diameter 70 cm covered a distance of 350 cm in 2 seconds. How many radians per second did it turn?

5 Juist antwoord

Antwoorddetails

To find the radians per second that the bicycle wheel turns, we need to find the angle in radians that the wheel turns in 1 second. The distance covered by the bicycle wheel in 1 second can be found by dividing the distance covered in 2 seconds by 2: \[\frac{350 \, \text{cm}}{2 \, \text{s}} = 175 \, \text{cm/s}\] To convert this to radians per second, we need to know the circumference of the wheel in centimeters. The circumference of the wheel is given by: \[\pi \times \text{diameter} = \pi \times 70 \, \text{cm} = 220 \pi \, \text{cm}\] So, in one revolution, the wheel covers a distance of 220π cm. The angle in radians that the wheel turns in 1 second can be found by dividing the distance covered in 1 second by the circumference of the wheel: \[\frac{175 \, \text{cm/s}}{220 \pi \, \text{cm/rev}} = \frac{5}{2\pi} \, \text{rev/s}\] To convert revolutions per second to radians per second, we multiply by 2π: \[\frac{5}{2\pi} \times 2\pi = 5 \, \text{rad/s}\] Therefore, the bicycle wheel turns at a rate of 5 radians per second. So the answer is (A) 5.

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Vectors

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Vraag 11 Verslag

If $16^{3x} = \frac{1}{4}(32^{x - 1})$, find the value of x.

$-1$ Juist antwoord

$\frac{-1}{3}$

$\frac{-3}{7}$

$\frac{-5}{19}$

Antwoorddetails

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Surds

Sets

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Vraag 12 Verslag

Find the axis of symmetry of the curve $y = x^{2} - 4x - 12$.

x = -2

y = -2

x = 2 Juist antwoord

y = 2

Antwoorddetails

To find the axis of symmetry of a quadratic equation in the form of $y = ax^2 + bx + c$, we need to use the formula: $$x = \frac{-b}{2a}$$ Comparing the given equation $y = x^2 - 4x - 12$ with the standard form of the quadratic equation, we have: $$a = 1, b = -4, c = -12$$ Substituting the values in the formula, we get: $$x = \frac{-(-4)}{2(1)} = 2$$ Therefore, the axis of symmetry of the curve is the vertical line passing through the point $(2,0)$ and the equation of this line is: $$x = 2$$ Hence, the correct answer is "x = 2".

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Vraag 13 Verslag

P, Q, R, S are points in a plane such that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).

(-6, -15)

(-6, 5)

(20, 5) Juist antwoord

(20, 15)

Antwoorddetails

Given that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j, and PS = xi + yj. Since PQ + QR + RS = PS, we have: (8i - 5j) + (5i + 7j) + (7i + 3j) = xi + yj Simplifying the left-hand side, we get: 20i + 5j = xi + yj Equating the corresponding components, we have: 20 = x and 5 = y Therefore, the coordinates of point S are (20, 5). Hence, the correct answer is (20, 5).

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Vectors

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Vraag 14 Verslag

Find the equation of the tangent to the curve $y = 4x^{2} - 12x + 7$ at point (2, -1).

y + 4x - 9 = 0

y - 4x - 9 = 0

y - 4x + 9 = 0 Juist antwoord

y + 4x + 9 = 0

Antwoorddetails

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Vraag 15 Verslag

The distance s metres of a particle from a fixed point at time t seconds is given by $s = 7 + pt^{3} + t^{2}$, where p is a constant. If the acceleration at t = 3 secs is $8 ms^{-2}$, find the value of p.

$\frac{1}{3}$ Juist antwoord

$\frac{4}{9}$

$\frac{5}{9}$

$1$

Antwoorddetails

Given, distance of a particle from a fixed point at time t seconds is given by, $$s = 7 + pt^{3} + t^{2}$$ We can find the acceleration of the particle by differentiating the distance equation twice with respect to time. $$\frac{d}{dt} s = \frac{d}{dt} (7 + pt^{3} + t^{2})$$ $$\Rightarrow \frac{ds}{dt} = 3pt^2 + 2t$$ $$\frac{d^{2}}{dt^{2}} s = \frac{d}{dt} (3pt^2 + 2t)$$ $$\Rightarrow \frac{d^{2}s}{dt^{2}} = 6pt + 2$$ We are given that the acceleration at t = 3 sec is $8 ms^{-2}$, so we can substitute the values in the second derivative of the distance equation to get, $$6p(3) + 2 = 8$$ $$\Rightarrow 18p = 6$$ $$\Rightarrow p = \frac{6}{18} = \frac{1}{3}$$ Hence, the value of p is $\frac{1}{3}$. Therefore, is correct.

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Polynomial Functions

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Vraag 16 Verslag

The third of geometric progression (G.P) is 10 and the sixth term is 80. Find the common ratio.

2 Juist antwoord

Antwoorddetails

We can use the formula for the nth term of a geometric progression to solve this problem. Let a be the first term and r be the common ratio, then the third term is ar^2 and the sixth term is ar^5. We are given that ar^2 = 10 and ar^5 = 80. Dividing the second equation by the first, we get: (ar^5)/(ar^2) = 80/10 Simplifying and canceling a, we get: r^3 = 8 Taking the cube root of both sides, we get: r = 2 Therefore, the common ratio is 2, and the answer is (a) 2.

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Sets

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Vraag 17 Verslag

Evaluate $\int_{1}^{2} \frac{4}{x^{3}} \mathrm {d} x$

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

$-\frac{15}{16}$

$\frac{15}{16}$

$1\frac{1}{2}$ Juist antwoord

Antwoorddetails

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Integration

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Vraag 18 Verslag

The coefficient of the 7th term in the binomial expansion of $(2 - \frac{x}{3})^{10}$ in ascending powers of x is

$\frac{560}{243}$

$\frac{841}{243}$

$\frac{1120}{243}$ Juist antwoord

$\frac{4481}{243}$

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Binomial Theorem

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Vraag 19 Verslag

The initial velocity of an object is $u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}$. If the acceleration of the object is $a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}$ and it moved for 3 seconds, find the final velocity.

$\begin{pmatrix} -14 \\ 15 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} -2 \\ 1 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$ Juist antwoord

$\begin{pmatrix} 14 \\ -9 \end{pmatrix} ms^{-1}$

Antwoorddetails

The final velocity of an object can be found using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we have: v = (-5, 3) + 3(3, -4) v = (-5, 3) + (9, -12) v = (4, -9) Therefore, the final velocity of the object is $\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$. The correct option is (C).

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Matrices And Linear Transformation

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Vraag 20 Verslag

Find the value of $\cos(60° + 45°)$ leaving your answer in surd form.

$\frac{6 + \sqrt{2}}{4}$

$\frac{3 + \sqrt{6}}{4}$

$\frac{\sqrt{2} - \sqrt{6}}{4}$ Juist antwoord

$\frac{3 - \sqrt{6}}{4}$

Antwoorddetails

Using the sum to product formula, we have: \begin{align*} \cos(60° + 45°) &= \cos 60° \cos 45° - \sin 60° \sin 45° \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{align*} Therefore, the correct answer is option C: $\frac{\sqrt{2} - \sqrt{6}}{4}$.

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Surds

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Vraag 21 Verslag

Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?

$\frac{1}{24}$ Juist antwoord

$\frac{1}{6}$

$\frac{2}{13}$

$\frac{1}{4}$

Antwoorddetails

There are 4 seats to be labeled, so Yomi can label the first seat with any of the 4 letters. After labeling the first seat, there are 3 letters left to choose from for the second seat, 2 letters for the third seat, and only 1 letter left for the last seat. Thus, there are a total of 4 x 3 x 2 x 1 = 24 possible ways to label the seats. Out of these 24 possible ways, only one arrangement is in alphabetical order (i.e., S, P, Q, R). Therefore, the probability that Yomi labeled the seats in alphabetical order is 1/24. Therefore, the correct option is (a) $\frac{1}{24}$.

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Probability

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Vraag 22 Verslag

A company took delivery of 12 vehicles made up of 7 buses and 5 saloon cars for two of its departments; Personnel and General Administration. If the Personnel department is to have at least 3 saloon cars, in how many ways can these vehicles be distributed equally between the departments?

350

455

462 Juist antwoord

571

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Permutation And Combinations

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Vraag 23 Verslag

If (x - 3) is a factor of $2x^{2} - 2x + p$, find the value of constant p.

-12 Juist antwoord

-6

Antwoorddetails

If (x-3) is a factor of the quadratic expression, then the expression can be factored into the form: $$(x - 3)(ax + b)$$ where a and b are constants. Multiplying out the brackets gives: $$2x^{2} - 2x + p = (x - 3)(ax + b) = ax^{2} + (b - 3a)x - 3b$$ Since the coefficients of the quadratic expression are equal to those of the factored expression, we can equate the corresponding coefficients to get a system of equations: $$a = 2$$ $$b - 3a = -2$$ $$-3b = p$$ Solving these equations simultaneously, we get: $$a = 2, b = 3a - 2 = 4, p = -3b = -12$$ Therefore, the constant p is -12.

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Polynomial Functions

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Vraag 24 Verslag

Find the direction cosines of the vector $4i - 3j$.

$\frac{9}{10}, \frac{27}{10}$

$\frac{17}{27}, -\frac{17}{27}$

$\frac{4}{5}, -\frac{3}{5}$ Juist antwoord

$\frac{4}{7}, \frac{-3}{7}$

Antwoorddetails

Direction cosines are the cosines of the angles that a vector makes with the positive x, y and z axes. Let's denote the vector by $\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} + 0\mathbf{k}$. Then the direction cosines of $\mathbf{v}$ are: \begin{align*} \cos\alpha &= \frac{4}{5} \\ \cos\beta &= \frac{-3}{5} \\ \cos\gamma &= 0 \end{align*} Therefore, the correct option is: - $\frac{4}{5}, -\frac{3}{5}$

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Vectors

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Vraag 25 Verslag

Two forces $F_{1} = (7i + 8j)N$ and $F_{2} = (3i + 4j)N$ act on a particle. Find the magnitude and direction of $F_{1} - F_{2}$.

$(4\sqrt{2} N, 000°)$

$(4\sqrt{2} N, 045°)$ Juist antwoord

$(4\sqrt{2} N, 090°)$

$(4\sqrt{2} N, 180°)$

Antwoorddetails

To find the magnitude and direction of $F_{1} - F_{2}$, we need to subtract the components of $F_{2}$ from $F_{1}$. \begin{align*} F_{1} - F_{2} &= (7i + 8j) - (3i + 4j)\\ &= 4i + 4j\\ \end{align*} The magnitude of $F_{1} - F_{2}$ is given by: \begin{align*} |F_{1} - F_{2}| &= \sqrt{(4)^2 + (4)^2}\\ &= 4\sqrt{2} N\\ \end{align*} The direction of $F_{1} - F_{2}$ is given by: \begin{align*} \theta &= \tan^{-1}\left(\frac{4j}{4i}\right)\\ &= \tan^{-1}(1)\\ &= 45°\\ \end{align*} Therefore, the magnitude and direction of $F_{1} - F_{2}$ are $(4\sqrt{2} N, 045°)$. Answer is correct.

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Statics

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Vraag 26 Verslag

If $\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = -2$, find the value of x.

-8 Juist antwoord

-4

Antwoorddetails

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Matrices And Linear Transformation

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Vraag 27 Verslag

Find the least value of n for which $^{3n}C_{2} > 0, n \in R$.

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{2}{3}$ Juist antwoord

Antwoorddetails

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Permutation And Combinations

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Vraag 28 Verslag

If $\sin x = -\sin 70°, 0° < x < 360°$, determine the two possible values of x.

110°, 250°

110°, 290°

200°, 250°

250°, 290° Juist antwoord

Antwoorddetails

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Trigonometry

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Vraag 29 Verslag

Given that $\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4$ and y = 6 when x = 3, find the equation for y.

$x^{3} - 4x - 9$ Juist antwoord

$x^{3} - 4x + 9$

$x^{3} + 4x - 9$

$x^{3} + 4x + 9$

Antwoorddetails

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Vraag 30 Verslag

If $\overrightarrow{OA} = 3i + 4j$ and $\overrightarrow{OB} = 5i - 6j $ where O is the origin and M is the midpoint of AB, find OM.

-2i - 10j

-2i + 2j

4i - j Juist antwoord

4i + j

Antwoorddetails

We can find the midpoint M of AB by using the midpoint formula: \[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\] Substituting the values we have: \[\overrightarrow{OM} = \frac{(3i + 4j) + (5i - 6j)}{2} = \frac{8i - 2j}{2} = 4i - j\] Therefore, the answer is 4i - j.

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Vectors

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Vraag 31 Verslag

If $\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = m\sqrt{2}$, where m is a constant. Find m.

$1\frac{1}{2}$

$1\frac{1}{4}$

$2\frac{1}{4}$ Juist antwoord

$2\frac{1}{2}$

Antwoorddetails

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Surds

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Vraag 32 Verslag

Given that $P = {x : \text{x is a factor of 6}}$ is the domain of $g(x) = x^{2} + 3x - 5$, find the range of x.

{-1, 5, 13}

{5, 13, 49}

{1, 2, 3, 6}

{-1, 5, 13, 49} Juist antwoord

Antwoorddetails

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Lees lesnotitie op Functions (WAEC)

Polynomial Functions

Functions

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Vraag 33 Verslag

If $y = 2(2x + \sqrt{x})^{2}$, find $\frac{\mathrm d y}{\mathrm d x}$.

$2\sqrt{x}(2x + \sqrt{2})$

$4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$ Juist antwoord

$4(2x + \sqrt{x})(2 + \sqrt{x})$

$8(2x + \sqrt{x})(2 + \sqrt{x})$

Antwoorddetails

To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation. Let's start by using the power rule of differentiation to find the derivative of $(2x + \sqrt{x})^{2}$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\] Next, we use the chain rule to differentiate $\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\] Substituting this into the previous equation, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\] Expanding this expression, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\] Simplifying further, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\] Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function: \[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\] Therefore, the correct answer is: $4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$.

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Vraag 34 Verslag

The roots of a quadratic equation are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$. Find its equation.

$x^{2} - 6x - 9 = 0$

$x^{2} - 6x + 6 = 0$ Juist antwoord

$x^{2} + 6x - 9 = 0$

$x^{2} + 6x + 6 = 0$

Antwoorddetails

The roots of a quadratic equation of the form $ax^{2} + bx + c = 0$ are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$, then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$ is $x^{2} - 6x + 6 = 0$. Therefore, the answer is.

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Surds

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Vraag 35 Verslag

A stone is thrown vertically upwards and its height at any time t seconds is $h = 45t - 9t^{2}$. Find the maximum height reached.

45. 25 m

45.50 m

56.00 m

56.25 m Juist antwoord

Antwoorddetails

The height of the stone at any time t seconds is given by $h = 45t - 9t^{2}$. To find the maximum height reached, we need to find the vertex of the parabolic function $h$, since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula $\frac{-b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic function. In this case, $a = -9$ and $b = 45$. Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute $t = 2.5$ into the equation for $h$: \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.

Lees lesnotitie op Differentiation (WAEC)

Differentiation

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Vraag 36 Verslag

In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?

0.10

0.30 Juist antwoord

0.60

0.70

Antwoorddetails

We can use the formula: P(A and B) = P(A) + P(B) - P(A or B) where A is the event of liking Science, B is the event of liking History, P(A and B) is the probability of selecting a pupil who likes both Science and History, and P(A or B) is the probability of selecting a pupil who likes either Science or History or both. We know that there are 50 pupils in the class, 35 like Science and 30 like History. This means that the maximum number of pupils who could like both Science and History is 30 (since 30 already like History). Therefore, we have: P(A and B) <= 30/50 = 0.6 Also, we know that P(A or B) = P(A) + P(B) - P(A and B) (as shown above). Therefore: P(A and B) = P(A) + P(B) - P(A or B) = 35/50 + 30/50 - P(A or B) = 65/50 - P(A or B) We also know that P(A or B) <= 1, since it is the probability of selecting a pupil who likes either Science or History or both. Therefore: P(A and B) >= 65/50 - 1 = 15/50 = 0.3 Thus, the probability of selecting a pupil who likes both Science and History is between 0.3 and 0.6. The only option that falls within this range is 0.30, so the answer is: - 0.30

Lees lesnotitie op Probability (WAEC)

Probability

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Vraag 37 Verslag

The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.

m + n

m + n - 2mn Juist antwoord

1 - mn

Antwoorddetails

The probability that only one of the husband and wife will be alive in 15 years can happen in two ways: either the husband is alive and the wife is not, or the wife is alive and the husband is not. The probability of the first case is (m)(1-n) = m - mn, and the probability of the second case is (n)(1-m) = n - mn. Therefore, the total probability that only one of them will be alive is (m - mn) + (n - mn) = m + n - 2mn. Hence, the correct answer is: m + n - 2mn.

Lees lesnotitie op Probability (WAEC)

Probability

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Vraag 38 Verslag

Simplify $\frac{\log_{5} 8}{\log_{5} \sqrt{8}}$.

-2

$\frac{-1}{2}$

$\frac{1}{2}$

2 Juist antwoord

Antwoorddetails

We can use the properties of logarithms to simplify this expression. Recall that $\log_{a} b^{c} = c\log_{a} b$ and $\log_{a} \sqrt{b} = \frac{1}{2}\log_{a} b$. So, \[\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8} = \frac{2\log_{5} 8}{\log_{5} 8} = \boxed{2}.\] Therefore, the answer is 2.

Lees lesnotitie op Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

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Vraag 39 Verslag

Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.

70.6° Juist antwoord

50.2°

39.8°

19.4°

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Lees lesnotitie op Matrices And Linear Transformation (WAEC)

Lees lesnotitie op Co-ordinate Geometry (WAEC)

Matrices And Linear Transformation

Co-ordinate Geometry

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Vraag 40 Verslag

Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.

$2\sqrt{61}$

$\sqrt{42}$

$2\sqrt{21}$ Juist antwoord

$2\sqrt{10}$

Antwoorddetails

Lees lesnotitie op Trigonometry (WAEC)

Trigonometry

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WAEC SSCE - Further Mathematics - 2010 (Objective)