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Goshia [24]
3 years ago
11

If f(x)=5x^4+3, then what is the remainder when f(x) is divided by x-1?

Mathematics
2 answers:
KatRina [158]3 years ago
3 0
8


This is actually fairly simple. If you take the x-1 and set it equal to 0 you get 1. And if you plug that value into the x you get 8, which is the remainder. This is called the remainder value theorem.
adell [148]3 years ago
3 0

Answer:

8

This is actually fairly simple. If you take the x-1 and set it equal to 0 you get 1. And if you plug that value into the x you get 8, which is the remainder. This is called the remainder value theorem.

Step-by-step explanation:

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6-5 solving rational equations 8/x+3=1/x+1
AfilCa [17]

Answer:

x = -5/7

Step-by-step explanation:

8/(x+3)=1/(x+1)

Using cross products

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3 years ago
The question is:
saul85 [17]
f(x) = \sqrt{x+1} \\ y = \sqrt{x+1} \ or \ y^2 = x+1 \ or \ x = y^2 - 1 \\ Therefore, \ f^{-1}(x)=x^2 - 1

(fof^{-1})(a) = f(f^{-1}(a))=f(a^2-1) \\ =\sqrt{a^2-1+1}=\sqrt{a^2}=a \\  \\ (f^{-1}of)(a)=f^{-1}(f(a))=f^{-1}(\sqrt{a+1})= \\ (\sqrt{a+1})^2-1 = a+1-1 = a

Therefore, (fof^{-1})(a) = (f^{-1}of)(a) = a

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