WILL GIVE BRAINLIST... do i do 50 cosine 90 here? not sure.

Mathematics

2 answers:

Setler79 [48]3 years ago

4 0

Answer:

x = 117.9°

Step-by-step explanation:

We require to find the size of the angle x

Given the law of cosines

c² = a² + b² - 2ab cosC

Rearrange to give cosC in terms of the sides

Add 2ab cosC to both sides

c² + 2ab cosC = a² + b² ( subtract c² from both sides )

2ab cosC = a² + b² - c² ( divide both sides by 2ab )

cosC = $\frac{a^2+b^2-c^2}{2ab}$

here a = 55, b = 50 and c = 90 and C = x

cosx = $\frac{55^2+50^2-90^2}{2(55)(50)}$

= $\frac{3025+2500-8100}{5500}$

= $\frac{-2575}{5500}$ ← - indicates angle is obtuse

x = $cos^{-1}$ ( $\frac{-2575}{5500}$ ) ≈ 117.9°

natulia [17]3 years ago

3 0

The answer should be : 117.9

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irga5000 [103]

Answer:

None of these.

Step-by-step explanation:

Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.

Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.

In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.

If we look at the first term x^3 and we divide it by 13 we get $\frac{x^3}{13}$ we cannot reduce it so it is not a fraction so 13 is not a factor of P