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tankabanditka [31]
4 years ago
10

Is this a A) Acute B)Equiangular C)obtuse D)right

Mathematics
2 answers:
vampirchik [111]4 years ago
8 0

The answer is A) Acute


dangina [55]4 years ago
7 0
An Acute angle because anythung 70° is acute
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Rewrite without parentheses.<br> -3a^2c^4(5c^3+4a-8)<br> Simplify your answer as much as possible.
yanalaym [24]

Answer:

=−15a2c7−12a3c4+24a2c4

Step-by-step explanation:

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4 years ago
The endpoints of EFare E(xE , yE) and F(xF , yF). What are the coordinates of the midpoint of EF?
jekas [21]
<span>The coordinates of the midpoint of EF is ((xE + xF)/2, (yE + yF)/2)
The third option is the correct answer.
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I think the Volume of the Square is 210.

Step-by-step explanation:

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3 years ago
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A group consisting of 10 children an adult went to a movie theater children tickets cost 5 each and adults tickets cost 8 each a
wlad13 [49]

9514 1404 393

Answer:

  6 children

Step-by-step explanation:

If all of the group were adults, the total cost of tickets would be 8×$10 = $80. The actual cost was $80 -62 = $18 less than that. Since each children's ticket saves $3 off the cost of an adult ticket, there were $18/$3 = 6 children's tickets.

There were 6 children in the group.

_____

If you want an equation, you can let c represent the number of children. Then the cost of tickets is ...

  8(10 -c) +5c = 62 . . . . 10-c is the number of adults

  -3c = -18 . . . . . . . subtract 80

  c = -18/-3 = 6 . . . divide by 3

3 0
3 years ago
Find both the number of combinations and the number of permutations for the given number of objects.
Nikitich [7]

Answer:

First, if we have a set of K elements, such that are ordered as:

{x₁, x₂, ...}

The total number of permutations for the K elements can be found in the next way.

For the first element in the set, we have K options.

For the second element in the set, we have (K - 1) options (because we already choose one)

For the third element we have (K - 2) options, and so on.

The total number of permutation is equal to the product between the numbers of options for each position's element, then the number of permutations for K elements is:

permutations = K*(K - 1)*(K - 2)*....*2*1 = K!

Now suppose that we have a set of N elements, and we want to make groups of K elements.

The total number of different combinations of K elements is given by the equation:

C(N, K) = \frac{N!}{(N - K)!*K!}

In this case we have 15 objects (then  N = 15) and we take 7 at the time (Then K = 7)

Where we need to take in account the number of combinations and also the permutations for each combination.

Then the total number of different sets is:

C(15*7)*7!

First, the total number of combinations will be:

C(15,7) = \frac{15!}{(15 - 7)!7!} = \frac{15!}{8!*7!}  = \frac{15*14*13*12*11*10*9}{7*6*5*4*3*2*1}  = 6,435

So we have 6,436 combinations, and each one of these combinations has 7! permutations.

permutations = 7! = 7*6*5*4*3*2*1 = 5,040

if we combine these we get:

Combinations*Permutations = 6,435*5,040 = 32,432,400

5 0
3 years ago
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