Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6
76\(^∘\)
53\(^∘\)
37\(^∘\)
14\(^∘\)
Correct answer is C
| tanθ | = | m1 - m2
1 + m1m2 |
y = 2x + 5
m1 = 2
2y = x - 6
| y | = | 1
2 |
x | - | 3 |
| m2 | = | 1
2 |
| tanθ | = | 2 - \(\frac{1}{2}\)
1+2(\(\frac{1}{2}\)) |
tanθ = \(\frac{3}{2}\) ÷ (1+1)
tanθ = \(\frac{3}{2}\) ÷ 2
| tanθ | = | 3
4 |
θ = \(tan^{-1} (\frac{3}{4})\)
θ = 36.87º
θ = 37º
1.63m
1.54m
1.52m
1.42m
Correct answer is B
for 20 students, mean = 1.67
| μ | = | ∑fx
f |
∑fx = μf
∑fx = 20 × 1.67 = 33.4
for group 2
∑fx = 16 × 1.50 = 24
for group 3
∑fx = 14 × 1.40 = 19.6
33.4 + 24 + 19.6 = 77
∑f = 20+16+14 = 50
| μ | = | 77
50 |
= | 1.54m |
\(\frac{1}{2}\) m/s
\(\frac{1}{3}\) m/s
2m/s
3m/s
Correct answer is A
m1u1 + m2u2 = (m1 + m2)v
m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s
18(4) + 6(-10) = (18+6)v
72 - 60 = 24v
12 = 24v
v = \(\frac{1}{2}\) m/s
\(\frac{-1}{2}, 8\)
\(\frac{1}{2}, -8\)
\(\frac{-1}{2}, -8\)
\(\frac{1}{2}, 8\)
Correct answer is A
U1 = x - 4
U2 = x + 2
U3 = 3x + 1
\(\frac{u_2}{u_1} = \frac{u_3}{u_2}\)
\(\frac{x+2}{x-4} = \frac{3x+1}{x+2}\)
(x+2)(x+2) = (x-4)(3x+1)
x\(^2\) + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x\(^2\) - 15x - 8 =0
2x\(^2\) + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0
x = (\(\frac{-1}{2}, 8\))
Find the coefficient of x\(^3\)y\(^2\) in the binomial expansion of (x-2y)\(^5\)
-80
10
40
90
Correct answer is C
x\(^3\)y\(^2\) in (x-2y)\(^5\)
n = 5, r = 3, p = x, q = -2y
5C\(_3\) * x\(^3\) -2y\(^2\)
5C\(_3\) = \(\frac{5!}{[5-3]!3!}\)
\(\frac{5*4*3!}{2! 3!}\) → \(\frac{5*4}{2}\)
5C\(_3\) = 10
: 5C\(_3\) * x\(^3\) -2y\(^2\) = 10 * x\(^3\) 4y\(^2\)
40x\(^3\)y\(^2\)
the coefficient is 40