Factorize completely: 6ax - 12by - 9ay + 8bx
(2a - 3b)(4x + 3y)
(3a + 4b)(2x - 3y)
(3a - 4b)(2x + 3y)
(2a + 3b)(4x -3y)
Correct answer is B
6ax - 12by - 9ay + 8bx
= 6ax - 9ay + 8bx - 12by
= 3a(2x - 3y) + 4b(2x - 3y)
= (3a + 4b)(2x - 3y)
If 2n = y, Find 2\(^{(2 + \frac{n}{3})}\)
4y\(^\frac{1}{3}\)
4y\(^-3\)
2y\(^\frac{1}{3}\)
2y\(^-3\)
Correct answer is A
If 2n = y,
then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\)
= 4 x (2n)\(^{\frac{1}{3}}\)
But y = 2n, hence
2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\)
= 4y\(^\frac{1}{3}\)
Given that logx 64 = 3, evaluate x log\(_2\)8
6
9
12
24
Correct answer is C
If logx 64 = 3, then \(64 = x^3\)
\(4^3 = x^3\)
Since the indices are equal,
x = 4
Hence, x log\(_2\)8
= 4(3.log\(_2\)2)
= where log\(_2\)2 = 1(1)
= 12 X 1
= 12
(x + 3)2
(x + 5)(x − 3)
(x − 5)2
(x − 5)(x + 3)
Correct answer is D
x2 − 2x − 15
(x2 − 5x) + (3x − 15)
x(x − 5)+ 3(x − 5)
(x − 5)(x + 3)
Given that A = {1, 5, 7} B = {3, 9, 12, 15} C = {2, 4, 6, 8} Find (A ∪ B) ∪ C
{1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}
{1, 2, 3, 5, 6, 8, 12, 15}
{2, 4, 5, 9, 12, 15}
{1, 5, 6, 7, 8, 9, 12, 15}
Correct answer is A
{A ∪ B ∪ C}
{A ∪ B} = {1, 3, 5, 7, 9, 12, 15}
∪ C={2,4,6,8}
{A ∪ B} ∪ C = {1, 3, 5, 7, 9, 12, 15} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}
{A ∪ B} ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}