Find the range of values of x for which 2x\(^2\) + 7x - 15 ≥ 0.
x ≤ -5 or x ≥ \(\frac{3}{2}\)
x ≥ -5 or x ≤\(\frac{3}{2}\)
-5 ≤ x ≤ \(\frac{3}{5}\)
\(\frac{3}{5}\) ≤ x ≤ -5
Correct answer is A
2x\(^2\) + 7x - 15 ≥ 0
2x\(^2\) -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0
x ≤ -5 or x ≥ \(\frac{3}{2}\)
Solve: 4sin\(^2\)θ + 1 = 2, where 0º < θ < 180º
60º 0r 120º
30º 0r 150º
30º 0r 120º
60º 0r 150º
Correct answer is B
4sin\(^2\)θ + 1 = 2
4sin\(^2\)θ = 2 - 1
4sin\(^2\)θ = 1
\(\sqrt sin^2θ\) = \(\sqrt \frac{1}{4}\)
sinθ = \(\frac{1}{2}\)
θ = \(sin^{-1} \frac{1}{2}\)
θ = 30º 0r 150º
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