g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4
find f {g (-3)}.
37
1
-3
-179
Correct answer is A
g(-3) = 2(-3) + 3 = -6 + 3 = -3
F(-3) = 3 (-3)\(^2\) - 2(-3) + 4
= 27 + 6 + 4 = 37
Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
\(\frac{7}{5}\)
\(\frac{6}{7}\)
\(\frac{-7}{6}\)
\(\frac{-6}{7}\)
Correct answer is A
(\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
3\(^{-2(x + 2)}\) = 3\(^{5(x - 2)}\);
-2(x+2) = 5(x - 2)
-2x -4 = 5x - 10
-7x = -6
x = \(\frac{6}{7}\)
Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
log \(\frac{2}{7}\)
log 2
log \(\frac{2}{9}\)
log 9
Correct answer is C
\(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
= log 2 + log 4 - log 36
= log \(\frac{8}{36}\)
= log \(\frac{2}{9}\)
\(\begin{pmatrix} 1 & 6 \\ 1 & 18 \end{pmatrix}\)
\(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 6 \\ -1 & -18 \end{pmatrix}\)
\(\begin{pmatrix} -1 & -6 \\ -1 & -18 \end{pmatrix}\)
Correct answer is B
3M = 3 \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\)
= \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\)
2N = 2 \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\)
= \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)
(3M - 2N) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)
For what range of values of x is x\(^2\) - 2x - 3 ≤ 0
{x: -1 ≤ x ≤ 3}
{x: -3 ≤ x ≤ 1}
{x: -3 ≤ x ≤ -1}
{x: 1 ≤ x ≤ 3}
Correct answer is A
Let x\(^2\) - 2x - 3 = 0;
(x+1)(x-3) = 0
The zeros of the function are x = -1 and 3
The solution set is { x: -1 ≤ x ≤ 3}
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