\(\frac{-1}{2}, 8\)
\(\frac{1}{2}, -8\)
\(\frac{-1}{2}, -8\)
\(\frac{1}{2}, 8\)
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
If g(x) = √(1-x\(^2\)), find the domain of g(x)
x < -1 or x > 1
x ≤ -1 or x ≥1
-1 ≤ x ≤ 1
-1 < x < 1
Correct answer is C
1 - x\(^2\) ≥ 0
-x\(^2\) ≥ -1
x\(^2\) ≤ 1
√x\(^2\) ≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | x | 9 | 2x | 1 |
5
6
7
8
Correct answer is C
7km has the highest frequency(14)
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | x | 9 | 2x | 1 |
If a hunter is selected at random, find the probability that the hunter covered at least 6km.
\(\frac{3}{5}\)
\(\frac{2}{5}\)
\(\frac{3}{8}\)
\(\frac{9}{40}\)
Correct answer is A
5+4+x+9+2x+1 = 40
19+3x = 40
3x = 21
x = 7
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | 7 | 9 | 14 | 1 |
The probability that the hunter covered at least 6km, means the hunter covered either 6km or 7km, or 8km.
24 hunters covered at least 6km
| 24
40 |
= | 3
5 |