Correct Answer: A. 1458

Correct Answer: C. 19/2

Explanation

The mean is the average of all the numbers in the set of data. Add all the numbers and divide by 6. (4+1+25+2+10+15)/6= 57/6= 19/2 The answer is: 19/2

Correct Answer: C. 150 m

Explanation

The acceleration is between the times 3 secs and 8 secs.

Distance = \(velocity \times time\)

= \(30 \times (8 - 3) = 30 \times 5\)

= \(150 m\)

Correct Answer: C. 60 years

Correct Answer: d. ₦196.00

Correct Answer: A. 324

Explanation

\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)

Let the power on x be t, so that the power on \(x^{-2}\) = 9 - t

\((x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3\)

\(3t = 3 + 18 = 21 \therefore t = 7\)

To obtain the coefficient of \(x^{3}\), we have

\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)

= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)

Explanation

Let C be the cost and n the number of tourists. then C = x + ny. Where x and y are constants

740 = x + 20y...........(1)

960 = x + 30y.............(2)

Solving (1) and (2) simultaneously will yield x = 300 and y = 22.

Then, C = 300 + 22n.

If we substitute 15 for n this implies that C = 300 + (22 x 15)= $630.00

Explanation

(a) \(\frac{5 + \sqrt{2}}{3 - \sqrt{2}} - \frac{5 - \sqrt{2}}{3 + \sqrt{2}}\)

= \(\frac{(5 + \sqrt{2})(3 + \sqrt{2}) - (5 - \sqrt{2})(3 - \sqrt{2})}{3^{2} - (\sqrt{2})^{2}}\)

= \(\frac{(15 + 5\sqrt{2} + 3\sqrt{2} + 2) - (15 - 5\sqrt{2} - 3\sqrt{2} + 2)}{9 - 2}\)

= \(\frac{17 + 8\sqrt{2} - 17 + 8\sqrt{2}}{7}\)

= \(\frac{0}{7} + \frac{16\sqrt{2}}{7}\)

\(\therefore a = 0 ; b = \frac{16}{7}\)

(b) \(3x - y - z = -2\)

\(x + 5y + 2z = 5\)

\(2x + 3y + z = 0\)

We write the set of equations in matrix form.

\(\begin{pmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)

We find the inverse of the matrix, say A,

\(|A| = \begin{vmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{vmatrix} \)

= \(3(5 - 6) + 1(1 - 4) - 1(3 - 10) = 1\)

Cofactors, C

\(C = \begin{pmatrix} -1 & 3 & -7 \\ -2 & 5 & -11 \\ 3 & 7 & 16 \end{pmatrix}\)

\(C^{T} = \begin{pmatrix} -1 & -2 & 3 \\ 3 & 5 & 7 \\ -7 & -11 & 16 \end{pmatrix}\)

Inverse \(\frac{C^{T}}{|A|} = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix}\)

\(x = \begin{pmatrix} -1 & -2 & -3 \\ 3 & 5 & 7 \\ -7 & -11 & -16 \end{pmatrix} \begin{pmatrix} -2 \\ 5 \\ 0 \end{pmatrix}\)

= \(\begin{pmatrix} 2 - 10 \\ -6 + 25 \\14 - 55 \end{pmatrix}\)

= \(\begin{pmatrix} -8 \\ 19 \\ -41 \end{pmatrix}\)

\(\therefore x = -8 ; y = 19 ; z = -41\)

Correct Answer: C. 6³/₈

Correct Answer: B. angle bisector of the two lines