Solve, correct to three significant figures, (0.3)\(^x\) = (0,5)\(^8\)
4.61
4.606
0.461
0.0130
Correct answer is A
(0,3)\(^x\) = (0.5)\(^8\)
xlog 0,3 = 9 log 0.5
x = \(\frac{8 \log 0.5}{\log 0.3}\)
= 4.606
70 kg ms\(^{-1}\)
55 kg ms\(^{-1}\)
50 kg ms\(^{-1}\)
35 kg ms\(^{-1}\)
Correct answer is A
Change in momentum
= F x t = 35N x 2
= 70kg ms\(^{-1}\)
Which of these inequalities is represented by the shaded portion of the graph?
2y + x - 3 < 0
2y - x - 3 < 0
2y - x + 3 < 0
2y + x +3 < 0
Correct answer is B
(0, 1.5), (-3, 0)
m = \(\frac{0 - 1.5}{-3, 0}\) = 0.5
0.5 = \(\frac{y - 1.5}{-3.0}\) = 0.5
y = 0.5x + 1.5
2y = x + 3
2y - x - 3 < 0
Find the constant term in the binomial expansion of (2x\(^2\) + \(\frac{1}{x^2}\))\(^4\)
10
12
24
42
Correct answer is B
6(2x\(^2\))\(^2\) (\(\frac{1}{x^2}\))\(^2\)
= 6 x 2
= 12
0 ms\(^{-2}\)
3 ms\(^{-2}\)
6 ms\(^{-2}\)
9 ms\(^{-2}\)
Correct answer is C
V = 3t\(^2\) - 6t
\(\frac{dx}{dt}\) = 6r - 6
at t = 4
\(\frac{dx}{dt}\) = 6 x 4 - 6 = 24 - 6
= 18
at t = 3
\(\frac{dy}{dt}\) = 6 x 4 - 6
= 24 - 6
= 18
at t = 3
\(\frac{dx}{dt}\) = 6 x 3 - 6 = 18 - 6
= 12
\(\frac{dy}{dt}\) = 18 - 12
= 6ms\(^{-2}\)
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