Calculate the distance between points L(-1, -6) and M(-3, -5)
√5
2√3
√20
√50
Correct answer is A
L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\)
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\)
D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\)
D = \(\sqrt{(-2)^2 + 1^2}\)
D = \(\sqrt{4 + 1}\)
D = \(\sqrt{5}\)
-\(\frac{1}{8}\)
\(\frac{3}{8}\)
\(\frac{5}{8}\)
\(\frac{1}{4}\)
Correct answer is D
-2p + r = 1.......(i)
2p + 3r = 0.......(ii)
r - 1 + 2p ........(iii)
2p + 3(1 + 2p) = 0
2p + 3(1 + 2p) = 0
2p + 3 + 6p = 0
3 - 8p = 0 \(\to\) 8p = 3
p = \(\frac{3}{8}\)
6 = 1 - 2 \(\frac{3}{8}\)
= 1 - \(\frac{6}{8}\)
\(\frac{8 - 6}{8}\) = \(\frac{2}{8}\)
= \(\frac{1}{4}\)
-3, -2
-5, -3
-2, -5
-3, -5
Correct answer is D
\(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\)
therefore, (x, y) = (-3, -5) respectively
What value of x will make the function x(4 - x) a maximum?
4
3
2
1
Correct answer is C
x(4 - x)
4x - x2
\(\frac{dy}{dx}\) = 4 - 2x
\(\frac{dy}{dx}\) = 0
2x = 4
x = \(\frac{4}{2}\)
= 2
Determine the value of x for which (x2 - 1)>0
x < -1 or x > 1
-1 < x < 1
x > 0
x < -1
Correct answer is A
x(x - 1) > 0 x < -1 or x > 1