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)
Find the remainder when X3 - 2X2 + 3X - 3 is divided by X2 + 1
2X - 1
X + 3
2X + 1
X - 3
Correct answer is A
X2 + 1 \(\frac{X - 2}{\sqrt{X^3 - 2X^2 + 3n - 3}}\)
= \(\frac {- 6X^3 + n}{-2X^2 + 2X - 3}\)
= \(\frac{(-2X^2 - 2)}{2X - 1}\)
Remainder is 2X - 1
Make R the subject of the formula if T = \(\frac {KR^2 + M}{3}\)
\(\sqrt\frac{3T - K}{M}\)
\(\sqrt\frac{3T - M}{K}\)
\(\sqrt\frac{3T + K}{M}\)
\(\sqrt\frac{3T - K}{M}\)
Correct answer is B
T = \(\frac{KR^2 + M}{3}\)
3T = KR2 + M
KR2 = 3T - M
R2 = \(\frac{3T - M}{K}\)
R = \(\sqrt\frac{3T - M}{K}\)