{y : y \(\in\) R, y \(\neq\) 0}
{y : y \(\in\) R, y \(\neq\) 1}
{y : y \(\in\) R, y \(\neq\) 5}
{y : y \(\in\) R, y \(\neq\) 7}
Correct answer is C
No explanation has been provided for this answer.
p = 5, q = 3
p = 5, q = -3
p = -5, q = -3
p = -5, q = 3
Correct answer is B
p + q = 2
p - q = 8
\(\overline{\frac{2p}{2} = \frac{10}{2}}\)
p = 5
from p + q = 2
5 + q = 2
q = 2 - 5
= -3
If \(\int^3_0(px^2 + 16)dx\) = 129. Find the value of p
9
8
7
6
Correct answer is A
\(\int^3_0(px^2 + 16)\) = 129
\(\frac{px^2 + 1}{0 + 1} + 16x|^3_0 = 129\)
\(\frac{px^3}{3} + 16x|^3_0 = 129\)
(\(\frac{p(3)^3}{3} + 16(3)\)) - 0 = 129
9p + 48 = 129
9p = 129 - 48
\(\frac{9p}{9} = \frac{81}{9}\)
p = 9
If cos x = -0.7133, find the values of x between 0\(^o\) and 360\(^o\)
44.5\(^o\) , 224.5\(^o\)
123.5\(^o\) , 190.5\(^o\)
135.5\(^o\) , 213.5\(^o\)
135.5\(^o\) , 224.5\(^o\)
Correct answer is D
cos x = -0.7133
x = cos \(^{-1}\)(0.7133)
= 44.496\(^o\)
x = 180 - 44.495\(^o\)
x = 135.5\(^o\)
and x = 180\(^o\) + 44.495\(^o\)
= 224.5\(^o\)
Find the inverse of \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 5 & 1 \\ -3 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)
\(\begin{pmatrix} -5 & 2 \\ -1 & 3 \end{pmatrix}\)
\(\begin{pmatrix} 5 & 1 \\ 2 & 3 \end{pmatrix}\)
Correct answer is B
Let A = \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)
|A| = (3 x 2 - 5 x 1)
= 6 - 5
= 1
A\(^{-1}\) = \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)