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Kaylis [27]
3 years ago
8

Can someone help me with this question?

Mathematics
1 answer:
Lena [83]3 years ago
8 0

Answer:

idk its hard to me too

Step-by-step explanation:

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How do I answer this question
soldi70 [24.7K]

The answer is C. 2.

This is because the part that is inside of the parentheses is equal to 4. Multiplying the fractions in the numerator is equal to 4^\frac{3}{2}.

\frac{4^\frac{3}{2}}{4^\frac{1}{2}}=4., using the exponential rule 3/2-1/2=2/2 or 1. 4^1=4.

Now, we reduced the fraction, so all that is left is (4)^\frac{1}{2}. This is equal to 2.

So your answer is 2.

I hope this helps :)

8 0
3 years ago
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aksik [14]
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5 0
3 years ago
Please help me with this math problem, urgent please
djverab [1.8K]

problem decoded dude

follow meh

6 0
3 years ago
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If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?
Keith_Richards [23]
\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}

so, the domain is all real numbers EXCEPT that one.
4 0
3 years ago
If y varies directly as x and y = 5.5 when x = 11, find x when y = 3.5
PtichkaEL [24]

Answer:

x=7

Step-by-step explanation:

y=mx

5.5=m×11

5.5÷11=m

m=1/2

y=mx

3.5=1/2×x

3.5÷1/2=x

x=7

4 0
3 years ago
Read 2 more answers
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