Solve the inequality -6(x + 3) \(\leq\) 4(x - 2)
x \(\leq\) 2
x \(\geq\) -1
x \(\geq\) -2
x \(\leq\) -1
Correct answer is B
-6(x + 3) \(\leq\) 4(x - 2)
-6(x +3) \(\leq\) 4(x - 2)
-6x -18 \(\leq\) 4x - 8
-18 + 8 \(\leq\) 4x +6x
-10 \(\leq\) 10x
10x \(\geq\) -10
x \(\geq\) -1
T varies inversely as the cube of R. When R = 3, T = \(\frac{2}{81}\), find T when R = 2
\(\frac{1}{18}\)
\(\frac{1}{12}\)
\(\frac{1}{24}\)
\(\frac{1}{6}\)
Correct answer is B
T \(\alpha \frac{1}{R^3}\)
T = \(\frac{k}{R^3}\)
k = TR3
= \(\frac{2}{81}\) x 33
= \(\frac{2}{81}\) x 27
dividing 81 by 27
k = \(\frac{2}{2}\)
therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)
When R = 2
T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)
= \(\frac{1}{12}\)
If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1\(\frac{7}{9}\)
20\(\frac{1}{4}\)
27
2\(\frac{1}{4}\)
36
Correct answer is D
x \(\alpha\sqrt y\)
x = k\(\sqrt y\)
81 = k\(\sqrt9\)
k = \(\frac{81}{3}\)
= 27
therefore, x = 27\(\sqrt y\)
y = 1\(\frac{7}{9}\) = \(\frac{16}{9}\)
x = 27 x \(\sqrt{\frac{16}{9}}\)
= 27 x \(\frac{4}{3}\)
dividing 27 by 3
= 9 x 4
= 36
Solve for x and y respectively in the simultaneous equations -2x - 5y = 3. x + 3y = 0
-3, -9
9, -3
-9,3
3, -9
Correct answer is C
-2x -5y = 3 x + 3y = 0 x = -3y -2 (-3y) - 5y = -3 6y - 5y = 3 y = 3 but, x = -3y x = -3(3) x = -9 therefore, x = -9, y = 3
Factorize completely 9y2 - 16X2
(3y - 2x)(3y + 4x)
(3y + 4x)(3y + 4x)
(3y + 2x)(3y - 4x)
(3y - 4x)(3y + 4x)
Correct answer is D
9y2 - 16x2
= 32y2 - 42x2
= (3y - 4x)(3y +4x)