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1. Find the total number of ratio parts:

The ratio of mangoes for Ade : Ben : Charity is 3:4:5.

Total parts in the ratio = 3 + 4 + 5 = 12 parts.

$\text{Total parts} = 3 + 4 + 5 = 12$

2. Determine the value of one ratio part:

Ben's share is 4 parts, and Ben received 20 mangoes.

So, 4 parts = 20 mangoes.

$1 part = \frac{20}{4} mangoes = 5 mangoes.$

$1 \text{ part} = \frac{20 \text{ mangoes}}{4 \text{ parts}} = 5 \text{ mangoes/part}$

3. Calculate the total number of mangoes shared out:

Multiply the total ratio parts by the value of one part:

$\text{Total mangoes} = \text{Total parts} \times \text{Value of 1 part}$

$\text{Total mangoes} = 12 \times 5 = 60 \text{ mangoes}$

According to the Factor Theorem, if a polynomial$P(x)$ is divisible by ,$(x-a)$ then $(x-a)$ is a factor of ,$P(x)$ which means that when$x=a$ is substituted into the polynomial, the result is zero, i.e., $P(a)=0$.

Explanation: The mode of a data set is the value that appears most frequently.

In the given data set:

⇒ 15 appears 2 times

⇒ 12 appears 3 times

⇒ 18 appears 2 times

⇒ 10 appears 1 time

The value 12 appears more often than any other value. Therefore, the mode is 12.

The formula for the $n^{th}$ term of an A.P. is $T_n = a + (n-1)d$ , where a is the first term, d is the common difference, and n is the term number.

Given $a = 5$ , $d = 3$ , and $n = 10$ .

$T_{10} = 5 + (10-1)3$

$T_{10} = 5 + (9)3$

$T_{10} = 5 + 27$

$T_{10} = 32$ .

For a $2 \times 2 matrix A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $ad - bc$ .

For matrix $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ ,

Determinant of $A = (2)(3) - (1)(4)$

$= 6 - 4$

$= 2$ .

The word REPORT has 6 letters.

Vowels: E, O (2 vowels)

Consonants: R, P, R, T (4 consonants). Note: There are two 'R's.

Positions: 1st, 2nd, 3rd, 4th, 5th, 6th.

Odd places: 1st, 3rd, 5th (3 places)

Even places: 2nd, 4th, 6th (3 places)

The condition "no vowel occupies an odd place" means that both vowels must be placed in the even positions.

1. Arrange the vowels in the even places:

There are 2 vowels (E, O). They are distinct. There are 3 even places (2nd, 4th, 6th).

The number of ways to arrange 2 distinct vowels in 3 distinct even places is a permutation:

$P(3, 2) = \frac{3!}{(3-2)!} = \frac{3!}{1!} = 3 \times 2 = 6$

2. Arrange the consonants in the remaining places:

There are 4 consonants (R, P, R, T). Two of these are 'R'.

After placing the 2 vowels, there are $6 - 2 = 4$ remaining places. These 4 places are the 3 odd places and the 1 unused even place.

The number of ways to arrange 4 consonants with 2 identical ('R') in 4 places is given by the formula for permutations with repetitions:

$\frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$

3. Total number of arrangements:

Multiply the number of ways to arrange the vowels by the number of ways to arrange the consonants:

$\text{Total ways} = 6 \times 12 = 72$

Explanation:

1. Median: The data set has 8 values (an even number). The median is the average of the two middle values, which are the 4th and 5th terms (x and 40).

⇒ $\text{Median} = \frac{x + 40}{2}$

⇒ Given, Median = 35.

$35 = \frac{x + 40}{2}$

$70 = x + 40$

$x = 70 - 40 = 30$

2. Mean: The mean is the sum of all values divided by the count.

⇒ $\text{Sum} = 20 + 25 + 30 + x + 40 + y + 55 + 60$

⇒ Substitute x=30:

$\text{Sum} = 20 + 25 + 30 + 30 + 40 + y + 55 + 60 = 260 + y$

⇒ Given, Mean = 40.

⇒ $40 = \frac{260 + y}{8}$

⇒ $320 = 260 + y$

⇒ $y = 320 - 260 = 60$

Therefore, x = 30 and y = 60.

Apply the power of a product rule $(xy)^n = x^n y^n$ and power of a power rule $(x^m)^n = x^{mn}$ :

$(a^3 b^{-2})^2 = (a^3)^2 (b^{-2})^2 = a^{3 \times 2} b^{-2 \times 2} = a^6 b^{-4}$ .

Now multiply by $a^{-1} b^5$ :

$a^6 b^{-4} \times a^{-1} b^5 = a^{6 + (-1)} b^{-4 + 5}$

$= a^{6 - 1} b^{1}$

$= a^5 b$ .

1. Convert volume from litres to $cm^3$ :

We know that 1 litre = 1000 $cm^3$ .

So, $385 litres = 385 \times 1000 \text{ cm}^3 = 385000 \text{ cm}^3$ .

$V = 385000 \text{ cm}^3$

2. Calculate the radius of the tank:

Diameter = 70 cm.

$Radius (r) = Diameter / 2 = 70 cm / 2 = 35 cm.$

$r = 35 \text{ cm}$

3. Use the formula for the volume of a cylinder to find height (h):

$Volume (V) = \pi r^2 h$

$V = \pi r^2 h$

Rearrange to solve for h:

$h = \frac{V}{\pi r^2}$

4. Substitute the values and calculate h:

Use $\pi = \frac{22}{7}.$

$h = \frac{385000}{\frac{22}{7} \times (35)^2}$

$h = \frac{385000}{\frac{22}{7} \times 1225}$

$h = \frac{385000}{22 \times \frac{1225}{7}}$

$h = \frac{385000}{22 \times 175}$

$h = \frac{385000}{3850}$

$h = 100 \text{ cm}$

Explanation:

⇒ Using the product rule of logarithms: $\log_b X + \log_b Y = \log_b (X \times Y)$

$\log_a 4 + \log_a 16 = \log_a (4 \times 16) = \log_a 64$

⇒ So the equation becomes:

$\log_a 64 = 3$

⇒ By definition of logarithm, if $\log_a X = Y$ , then $a^Y = X$ .

⇒ So, $a^3 = 64.$

⇒ To find a, take the cube root of 64:

⇒ $a = \sqrt[3]{64}$

⇒ Since $4 \times 4 \times 4 = 64$ ,

⇒ a = 4

The permutation formula is $nP_r = \frac{n\!}{(n-r)\!}$ .

The given ratio can be written as:

$\frac{12P_{k+1}}{11P_{k+1}} = \frac{12}{5}$

Expand the permutation terms:

$\frac{\frac{12!}{(12 - (k+1))!}}{\frac{11!}{(11 - (k+1))!}} = \frac{12}{5}$

Simplify the denominators:

$\frac{\frac{12!}{(11 - k)!}}{\frac{11!}{ (10 - k)!}} = \frac{12}{5}$

Rewrite the division as multiplication by the reciprocal:

$\frac{12!}{(11 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Use the property $n\! = n \times (n-1)\!$ :

$12\! = 12 \times 11\!$

$(11 - k)\! = (11 - k) \times (10 - k)\!$

Substitute these into the equation:

$\frac{12 \times 11!}{(11 - k) \times (10 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Cancel out $11\! and (10 - k)\!$ :

$\frac{12}{11 - k} = \frac{12}{5}$

Since the numerators are equal (12 on both sides), the denominators must also be equal:

11 - k = 5

Solve for k:

k = 11 - 5

k = 6

We can use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$ .

$(\frac{3}{5})^2 + \cos^2 \theta = 1$

$\frac{9}{25} + \cos^2 \theta = 1$

$\cos^2 \theta = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$

$\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$ (since $\theta$ is acute, $\cos \theta$ is positive).

Now, $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Given the points (-2, 5) and (6, -3):

$x_1 = -2, y_1 = 5$

$x_2 = 6, y_2 = -3$

Calculate the x-coordinate of the midpoint:

$x_{\text{mid}} = \frac{-2 + 6}{2} = \frac{4}{2} = 2$

Calculate the y-coordinate of the midpoint:

$y_{\text{mid}} = \frac{5 + (-3)}{2} = \frac{2}{2} = 1$

The coordinates of the midpoint are (2, 1).

Let the length be L = (x+5) cm.

Let the width be W = x cm.

Perimeter P = 2(L + W)

$30 = 2((x+5) + x)$

$30 = 2(2x + 5)$

$15 = 2x + 5$

$15 - 5 = 2x$

$10 = 2x$

$x = 5 \text{ cm}$

1. Find the union of B and C $(B \cup C)$ :

$B = \{b, c, d, e\}$

$C = \{a, c, e, g\}$

$B \cup C = \{a, b, c, d, e, g\}$

2. Find the intersection of $A and (B \cup C) (A \cap (B \cup C))$ :

$A = \{a, b, c\}$

$B \cup C = \{a, b, c, d, e, g\}$

$A \cap (B \cup C) = \{a, b, c\}$

To find $\frac{dy}{dx}$ for the implicit function $x^3 + y^3 = 5$ , we use implicit differentiation with respect to $x$ .

Differentiate each term with respect to x:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5)$

Apply the power rule for $x^3$ and the chain rule for $y^3$ :

$3x^2 + 3y^2 \frac{dy}{dx} = 0$

Rearrange the equation to solve for $\frac{dy}{dx}$ :

$3y^2 \frac{dy}{dx} = -3x^2$

Divide both sides by 3y^2:

$\frac{dy}{dx} = \frac{-3x^2}{3y^2}$

Simplify the expression:

$\frac{dy}{dx} = -\frac{x^2}{y^2}$

Explanation:

⇒ To differentiate $y = 3\cos(5x)$ , we use the chain rule.

⇒ Let u = 5x. Then $\frac{du}{dx} = 5$ .

⇒ The function becomes $y = 3\cos(u)$ .

⇒ Differentiate y with respect to u:

$\frac{dy}{du} = 3(-\sin(u)) = -3\sin(u)$ .

⇒ By the chain rule,

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ .

$\frac{dy}{dx} = (-3\sin(u)) \times 5 = -15\sin(u)$

⇒ Substitute u = 5x back: $\frac{dy}{dx} = -15\sin(5x)$ .

1. Find the reference angle:

First, find the acute angle (reference angle, let's call it $\beta$ ) for which $\cos \beta = \frac{\sqrt{3}}{2}$ .

$\cos \beta = \frac{\sqrt{3}}{2} \implies \beta = 30^\circ$

2. Determine the quadrants:

Since $\cos \alpha$ is negative $(-\frac{\sqrt{3}}{2})$ , $\alpha$ must lie in the quadrants where cosine is negative. These are the second (Q2) and third (Q3) quadrants.

3. Calculate $\alpha$ in Q2:

In the second quadrant, $\alpha$ $= 180^\circ - \beta$ .

$\alpha_1 = 180^\circ - 30^\circ = 150^\circ$

4. Calculate $\alpha$ in Q3:**

In the third quadrant, $\alpha = 180^\circ + \beta$ .

$\alpha_2 = 180^\circ + 30^\circ = 210^\circ$

Therefore, the values of $\alpha$ for which $\cos \alpha = -\frac{\sqrt{3}}{2}$ in the given range are $150^\circ$ and $210^\circ.$

According to the Remainder Theorem:

When $P(x)$ is divided by $(x-2)$ , the remainder is $P(2) = 7$ .

When $P(x)$ is divided by $(x+1)$ , the remainder is $P(-1) = 1$ .

Let the remainder when $P(x)$ is divided by $(x-2)(x+1)$ be $Ax + B$ , since the divisor is quadratic.

So, $P(x) = Q(x)(x-2)(x+1) + Ax + B$ .

Using the values from the Remainder Theorem:

For $x=2$ : $P(2) = A(2) + B \Rightarrow 2A + B = 7$ (Equation 1)

For $x=-1$ : $P(-1) = A(-1) + B \Rightarrow -A + B = 1$ (Equation 2)

Subtract Equation 2 from Equation 1:

$(2A + B) - (-A + B) = 7 - 1$

$2A + B + A - B = 6$

$3A = 6$

$A = 2$ .

Substitute $A=2$ into Equation 2:

$-2 + B = 1$

$B = 1 + 2 = 3$ .

So, the remainder is $Ax + B = 2x + 3$ .

Explanation:

1. Calculate the Loss:

$\text{Loss} = \text{Cost Price} - \text{Selling Price}$

$\text{Loss} = \text{₦}80,000 - \text{₦}64,000 = \text{₦}16,000$

2. Calculate Percentage Loss:

$\text{Percentage Loss} = \frac{\text{Loss}}{\text{Cost Price}} \times 100\%$

$\text{Percentage Loss} = \frac{\text{₦}16,000}{\text{₦}80,000} \times 100\%$

$\text{Percentage Loss} = \frac{16}{80} \times 100\%$

$\text{Percentage Loss} = \frac{1}{5} \times 100\%$

$\text{Percentage Loss} = 20\%$