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Len [333]
3 years ago
6

Can anyone tell me how to solve 8-(4c-5)=11?

Mathematics
2 answers:
muminat3 years ago
8 0
8-(4c-5)=11
8-4c+5=11
13-4c=11
-4c=-2
C=1/2
umka2103 [35]3 years ago
6 0

Answer:

all work is pictured/shown

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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

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