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gayaneshka [121]
3 years ago
8

How do I solve this? Please show work, thank you.

Mathematics
1 answer:
irina [24]3 years ago
7 0

Answer:

a   x sqrt(7) - 49sqrt(x)

Step-by-step explanation:

sqrt(7x) [ sqrt(x) - 7sqrt(7)]

Distribute

sqrt(7x) *sqrt(x) - sqrt(7x)*7sqrt(7)

We know that sqrt(a) sqrt(b)= sqrt(ab)

sqrt(7x^2)  - 7sqrt(7^2 *x)

Now lets separate out the perfect squares

sqrt(7) *sqrt(x^2) - 7sqrt(7^2)*sqrt(x)

x sqrt(7) - 7*7sqrt(x)

x sqrt(7) - 49sqrt(x)

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The hypotenuse and one of the legs of a right triangle form an angle that has a cosine of 2 √2 .
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Quick answer I don't think this has an answer.

If you take the cos-1(2 sqrt(2)) your calculator should have a fit. Let's check that out. Mine certainly does. So there is something wrong with the question. If there is something to add in please do it and I will it least put an answer in the comments. As it stands, nothing will work. 

If you put your calculator in radians, you will get an answer but it will not be anything resembling the choices you've listed.

If you meant sqrt(2) / 2 that would give 45o. Put it in your calculator like this 2 ^ 0.5 divided by 2 = 0.707
Cos - 1 (0.707) = 45



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