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antoniya [11.8K]
3 years ago
9

Algebra PLS HELP!!!!!!!!!!!!!

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

C - 53 Degrees

Step-by-step explanation:

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If the geometric mean of a and 34 is 6 sqrt(17) find the value of a
Alborosie

Using the geometric mean concept, it is found that the value of a is 18.

----------------------

The geometric mean, of a data-set of n elements, (n_1, n_2, ..., n_n), is given by:

G = \sqrt[n]{n_1 \times n_2 \times ... \times n_n}

That is, the nth root of the multiplication of all elements.

----------------------

In this question:

  • Two elements(n = 2), a and 34.
  • G = 6\sqrt{17}

Thus:

\sqrt{34a} = 6\sqrt{17}

We find the square of each side, so:

(\sqrt{34a})^2 = (6\sqrt{17})^2

34a = 36\times17

Simplifying both sides by 17:

2a = 36

a = \frac{36}{2}

a = 18

The value of a is 18.

A similar example is given at brainly.com/question/15010240

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3 years ago
Cameron is trying to find the equation of a line that
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Yes it does I went on Desmos graphing Calculator and put in the equation and got those answers
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3 years ago
I need help. Whats 50(3)^20?
Alex

Answer:

174339220050

Step-by-step explanation:

50(3^20)

=(50)(3486784401)

=174339220050

8 0
3 years ago
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What is the numerator in the fraction 64/72
Elodia [21]
The correct answer is 64
4 0
2 years ago
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So I have to compute the following but i don't know what they mean any help?
Gelneren [198K]

Answer:

Step-by-step explanation:

This is a Combination (as in permutation vs combination) question the symbol (n r) refers to "n choose r". This is sometimes written as nCr

i.e the question is asking you to find how many combinations each will yield when you chose r items from n item without repetition and order does not matter.

I will only do the first question for you and you can just follow the same steps to solve the rest of the questions.

Recall that

nCr=\frac{n!}{(r!)(n-r)!}

Consider question a) we are given (5 1) or ₅C₁

we can see that n = 5 and r = 1

If we substitute this into the formula:

₅C₁ = (5!) / [ (1!)(5- 1)!]

= (5!) / [ (5- 1)!]

= (5!) / (4!)

= (5·4·3·2·1) / (4·3·2·1)

= 5

hence ₅C₁ = 5

7 0
3 years ago
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