A set containing
elements has a power set containing
elements; here,
so
.
The elements of the power set are all possible combinations of up to 5 of the total 5 elements to choose from:
0 choices (1):
(since
)
1choice (5):
2 choices (10):
3 choices (10):
and so on
4 choices (5);
and so on
5 choices (1):
(1 + 5 + 10 + 10 + 5 + 1 = 32)
Answer:
13.8h
Step-by-step explanation:
Answer:
x = 15
Step-by-step explanation:
Because PQ is parallel to JL,
Substituting, we have
5(x-6) = 3x
5x - 30 = 3x
2x = 30
x = 15
Three unknowns requires three equations to solve . . .
1) n + d + q = 30
2) d = 2n
3) (0.05)n + (0.10)d + (0.25)q = 5.50
* plug 2) into 1) and solve for q
n + 2n + q = 30
3n + q = 30
q = 30 - 3n
* now we have variables d and q expressed in terms of n, so we plug those equations into 3) and solve for n . . .
0.05n + 0.10(2n) + 0.25(30 - 3n) = 5.50
0.05n + 0.20n + 7.50 - 0.75n = 5.50
(0.05 + 0.20 - 0.75)n + 7.50 = 5.50
-0.50n + 7.50 = 5.50
-0.50n = 5.50 - 7.50
-0.50n = -2.00
n = -2.00/-0.50 = 4
d = 2n = 2(4) = 8
q = 30 - 3n = 30 - 3(4) = 30 - 12 = 18
The answer is . . .
n = 4 nickels
d = 8 dimes
q = 18 quarters
. . . for a total of (4 + 8 + 18) = 30 coins
. . . 8 dimes is twice 4 nickels
. . . $0.05(4) + $0.10(8) + $0.25(18) = $5.50
Answer:
5.36
Step-by-step explanation:
Given that:
<BAD = <CAE, therefore, BD = EC
Let's take x to be the length of BD = EC
BD + DE + EC = BC
BC = 20,
BD = EC = x
DE ≈ 9.28
Thus,
x + 9.28 + x = 20
x + x + 9.28 = 20
2x + 9.28 = 20
Subtract 9.28 from both sides
2x + 9.28 - 9.28 = 20 - 9.28
2x = 10.72
Divided both sides by 2 to solve for x
BD ≈ 5.36