Correct Answer: B. 27

Explanation

\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)

Correct Answer: C. 3 + √6

Explanation

\(\frac{3}{3 - √3}\)

Rationalize

\(= \frac{3}{3 - √6} \times \frac{3 + √6}{3 + √6}\)

\(=\frac{3(3 + √6)}{(3 - √6)(3 + √6)}\)

\(=\frac{3(3 + √6)}{9 - 6}=\frac{3(3 + √6)}{3}\)

∴ 3 + √6

Explanation

(a) \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\)

Let \(u^{2} = 5 - x^{2}\).

\(2u \mathrm {d} u = - 2x \mathrm {d} x \implies u \mathrm {d} u = - x \mathrm {d} x\)

\(\int \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x = \int \frac{- u}{\sqrt{u^{2}}} \mathrm {d} u\)

= \(- \int \mathrm {d} u \)

= \(-[5 - x^{2}]_{1} ^{2}\)

= \((- (5 - 2^{2}) - (- (5 - 1^{2})))\)

= \(- 1 + 4 = 3\)

(b)(i) \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}\)

= \(2(-3 + 4) + 3(0 + 2) + 1(0 - 1) = 7\)

(ii) \(2x - 3y + z = 10\)

\(0x + y - 2z = -7\)

\(x + 2y - 3z = -9\)

Let A be the matrix of coefficients.

\(A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{pmatrix}\)

We write the set of equations in matrix form, i.e Ax = B.

\(\begin{pmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix}\).

We find the inverse of A, A\(^{-1}\).

Determinant of A = 7.

Co-factors:

\(A_{11} = + \begin{vmatrix} 1 & -2 \\ 2 & -3 \end{vmatrix} = +(-3 + 4) = 1\)

\(A_{12} = - \begin{vmatrix} 0 & -2 \\ 1 & 2 \end{vmatrix} = -(0 + 2) = -2\)

\(A_{13} = + \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = +(0 - 1) = -1\)

\(A_{21} = - \begin{vmatrix} -3 & 1 \\ 2 & -3 \end{vmatrix} = -(9 - 2) = -7\)

\(A_{22} = + \begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} = +(-6 - 1) = -7\)

\(A_{23} = - \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = -(4 - (-3)) = -7\)

\(A_{31} = + \begin{vmatrix} -3 & 1 \\ 1 & -2 \end{vmatrix} = +(6 - 1) = 5\)

\(A_{32} = - \begin{vmatrix} 2 & 1 \\ 0 & -2 \end{vmatrix} = -(-4 - 0) = 4\)

\(A_{33} = + \begin{vmatrix} 2 & -3 \\ 0 & 1 \end{vmatrix} = +(2 - 0) = 2\)

\(\therefore C = \begin{pmatrix} 1 & -2 & -1 \\ -7 & -7 & -7 \\ 5 & 4 & 2 \end{pmatrix}\)

\(\text{adj A} = C^{T} = \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(A^{-1} = \frac{\text{adj A}}{|A|} = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(x = A^{-1} b = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix} \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix}\)

= \(\frac{1}{7} \begin{pmatrix} 14 \\ -7 \\ 21 \end{pmatrix}\)

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

\(x = 2; y = -1 ; z = 3\)

Correct Answer: B. \(\text{1 and }\frac{-1}{3}\)

Explanation

For equal roots, \(b^{2} - 4ac = 0\)

From the equation, \(a = P, b = (P+1), c = P\)

\((P+1)^{2} - 4(P)(P) = P^{2} + 2P + 1 - 4P^{2} = 0\)

\(-3P^{2} + 2P + 1 = 0 \implies 3P^{2} - 2P - 1 = 0\)

\(3P^{2} - 3P + P - 1 = 0\)

\(3P(P - 1) + 1(P - 1) = 0\)

\(P = \text{1 or }\frac{-1}{3}\)

Correct Answer: B. \(-2 ≤ x ≤ 3\)

Explanation

\(x^2 - x - 4 ≤ 2\)

Subtract two from both sides to rewrite it in the quadratic standard form:

= \(x^2 - x - 4 - 2 ≤ 2 - 2\)

= \(x^2 - x - 6 ≤ 0\)

Now set it = 0 and factor and solve like normal.

= \(x^2 - x\) - 6=0

= \((x - 3)(x + 2)\)=0

\(x\) + 2 = 0 or \(x\) - 3 = 0

\(x\) = -2 or \(x\) = 3

So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, \(x\) = 0) and evaluate the original inequality.

= \(x2 - x - 4 ≤ 2\)

= \((0)^2 - (0) - 4 ≤ 2\)

= \(0 - 0 - 4 ≤ 2\)

\(−4 ≤ 2\)

Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.

∴ \(-2 ≤ x ≤ 3.\)

Correct Answer: B. \(-\frac{\sqrt{3}}{2}\)

Explanation

cos150° = -cos30°

= \(-\frac{\sqrt{3}}{2}\)

Correct Answer: A. 1

Explanation

Using the two - point from

\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)

\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)

\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)

\(\frac{-12(y -2)}{-4}\) = x - 4

3(y -2) = x -4

3y - 6 = x - 4

3y = x - 4 + 6

3y = x + 2...

By comparing the equations;

3y = px + , p = 1

Correct Answer: D. 6cm

Explanation

|AB| = 12 cm; |DC| = 9 cm; |AE| = 8 cm.

\(\Delta\) ABE and \(\Delta\) EDC are similar triangles. Hence,

\(\frac{|AB|}{|DC|} = \frac{|AE|}{|EC|}\)

\(\frac{12}{9} = \frac{8}{|EC|}\)

\(\therefore |EC| = \frac{9 \times 8}{12}\)

= 6 cm

Correct Answer: D. 94

Explanation

The equation to find the average of his five grades would look like (90+100+85+97+x)/5=94. You must solve for x, so the first step is to multiple both sides by 5 to get 90+100+85+97+x=470. Then you subtract all of the other grades from the total of 470 to get a final answer of 98 for the grade of the final exam

Correct Answer: D. 140°