Answer would be d- FB and DE
We would have the following sample space:
(1, 1), (1, 2), (1, 3), (1, 4)
(2, 1), (2, 2), (2, 3), (2, 4)
(3, 1), (3, 2), (3, 3), (3, 4)
(4, 1), (4, 2), (4, 3), (4, 4)
Those give us these sums:
2, 3, 4, 5
3, 4, 5, 6
4, 5, 6, 7
5, 6, 7, 8
P(sum of 2) = 1/16 =0.0625
P(sum of 3) = 2/16 = 0.125
P(sum of 4) = 3/16 = 0.1875
P(sum of 5) = 4/16 = 0.25
P(sum of 6) = 3/16 = 0.1875
P(sum of 7) = 2/16 = 0.125
P(sum of 8) = 1/16 = 0.0625
Answer:
(b) -m, m + 3
Step-by-step explanation:
x² − 3x − m(m + 3) = 0
x² − 3x = m(m + 3)
x² − 3x + 9/4 = m(m + 3) + 9/4
(x − 3/2)² = m(m + 3) + 9/4
(x − 3/2)² = m² + 3m + 9/4
(x − 3/2)² = (m + 3/2)²
x − 3/2 = ±(m + 3/2)
x − 3/2 = m + 3/2, -m − 3/2
x = m + 3, -m
Answer:
all real numbers except ±5.
Step-by-step explanation:
Each function (f(x), g(x)) has a domain that is all real numbers. Their quotient (f(x)/g(x)) must exclude values that make g(x) = 0. The quotient is undefined when the denominator is zero. Those excluded values are x = ±5.
The domain of (f/g)(x) is all real numbers except ±5.
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Of course you recognize x^2 -25 = 0 has solutions x = ±√25 = ±5. You can get there two ways:
- add 25 and take the square root: x^2=25; x=±√25.
- factor the difference of squares and set the factors to zero: (x-5)(x+5)=0 has solutions x-5=0 and x+5=0, that is, x = ±5.
Answer:
The answer to your question is the third choice
Step-by-step explanation:
To solve this problem, just multiply 6 by each of the numbers of the matrix and simplify.
6 | 4 -2 1 | = | (6 x 4) (6 x -2) (6 x 1) |
| 7 3 0| | (6 x 7) (6 x 3) (6 x 0) |
= | 24 -12 6 |
| 42 18 0 |
The first choice is wrong because only multiply by six the first row.
The second and fourth rows are incomplete.
