If log 5(\(\frac{125x^3}{\sqrt[ 3 ] {y}}\) is expressed in the values of p, q and k respectively
3, \(\frac{-1}{3}\), 5
\(\frac{-1}{3}\), 3, 5
3, \(\frac{-1}{3}\), 3
3, \(\frac{-1}{3}\), 3
Correct answer is D
log\(_5\) (\(\frac{125x^3}{\sqrt[3] {y}}\))
= \(\log_5 125 x^3 - \log _1 x^3 - log_5 y\frac{1}{3}\)
= \(3 log_5 5 + 3 log_5 x - \frac{1}{3} log _5 y\)
= 3, - \(\frac{1}{3}\), 3
If the sum of the roots of 2x\(^2\) + 5mx + n = 0 is 5, find the value of m
-2.5
-2.0
2.0
2.5
Correct answer is B
Sum of roots = \(\frac{-a}{b}\)
= \(\frac{-5m}{2}\) = 5
\(\frac{-5m}{m} = \frac{10}{-5}\)
m = -2
\(\frac{1}{5}\)(-3i - 4j)
\(\frac{1}{5}\)(-3i + 4j)
\(\frac{1}{5}\)(3i - 4j)
\(\frac{1}{5}\)(3i + 4j)
Correct answer is C
Resultant
(- 2i - 3j) + (5i - j)
= 3i - 4j
Unit vector
= \(\frac{3i - 4j}{|3i - 4j|} = \frac{3i - 4j}{\sqrt{3i } + (-4)}\)
= \(\frac{3i - 4j}{\sqrt{25}}\)
= \(\frac{3i - 4j}{5}\)
\(\frac{4}{3}\)
\(\frac{3}{4}\)
\(\frac{-3}{4}\)
\(\frac{-4}{3}\)
Correct answer is B
Gradient = \(\frac{-4 - 4}{-3 - 3}\)
= \(\frac{-8}{6}\)
= \(\frac{4}{3}\)
Normal = -(\(\frac{1}{\frac{1}{4}}\))
= \(\frac{-3}{4}\)
220
222
223
225
Correct answer is D
s = 120t - 16t\(^2\) - 16t\(^2\)
\(\frac{ds}{dt} = 120 - 32t\)
120 - 32t = 0
\(\frac{32t}{32}\) = \(\frac{120}{32}\)
t = 3.75
s = 120(3.75) - 16(3.75)\(^2\)
= 450 - 225
= 225