The value of y for which \(\frac{1}{5}y + \frac{1}{5} < \frac{1}{2}y + \frac{2}{5}\) is
\(y > \frac{2}{3}\)
\(y < \frac{2}{3}\)
\(y > -\frac{2}{3}\)
\(y < -\frac{2}{3}\)
Correct answer is C
\(\frac{1}{5}y + \frac{1}{5} < \frac{1}{2}y + \frac{2}{5}\)
Collect like terms
\(\frac{y}{5} - \frac{y}{2} < \frac{2}{5} - \frac{1}{5}\)
\(\frac{2y - 5y}{10} < \frac{2 - 1}{5}\)
\(\frac{-3y}{10} < \frac{1}{5}\)
\(y > \frac{-2}{3}\)
U is inversely proportional to the cube of V and U = 81 when V = 2. Find U when V = 3
24
27
32
36
Correct answer is A
U \(\propto \frac{1}{V^3}\)
U = \(\frac{k}{V^3}\)
k = UV\(^3\)
k = 81 x 2\(^3\) = 81 x 8
When V = 3,
U = \(\frac{k}{V^3}\)
U = \(\frac{81 \times 8}{3^3}\)
U = \(\frac{81 \times 8}{27}\) = 24
If y varies directly as \(\sqrt{n}\) and y = 4 when n = 4, find y when n = 1\(\frac{7}{9}\)
\(\sqrt{17}\)
\(\frac{4}{3}\)
\(\frac{8}{3}\)
\(\frac{2}{3}\)
Correct answer is C
y \(\propto \sqrt{n}\)
y = k\(\sqrt{n}\)
when y = 4, n = 4
4 = k\(\sqrt{4}\)
4 = 2k
k = 2
Therefore,
y = 2\(\sqrt{n}\)
y = 2\(\sqrt{\frac{16}{9}}\)
y = 2\((\frac{4}{3})\)
y = \(\frac{8}{3}\)
Solve for x and y in the equations below
x2 - y2 = 4
x + y = 2
x = 0, y = -2
x = 0, y = 2
x = 2, y = 0
x = -2, y = 0
Correct answer is C
x2 - y2 = 4 .... (1)
x + y = 2 .... (2)
Simplify eqn (1)
(x + y)(x - y) = 4
From eqn (2)
x + y = 2 so substitute it into simplified eqn (1), we have
2 (x - y) = 4
therefore,
x - y = 2 ... (1)
x + y = 2
---------
2x = 4
---------
x = 2, when y = 0
Find the remainder when 2x3 - 11x2 + 8x - 1 is divided by x + 3
-871
-781
-187
-178
Correct answer is D
Hence f(x) = 2x3 - 11x2 + 8x - 1
f(-3) = 2(-3)3 - 11(-3)2 + 8(-3) - 1
= 2(-27) - 11(9) + 8(-3) - 1
= -54 - 99 - 24 - 1
= -178