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3 years ago

Factor completely w^4-2401

2 answers:

6 0

$w^4-2401=(w^2)^2-49^2=(w^2-49)(w^2+49)=(w^2-7^2)(w^2+49)\\\\=(w-7)(w+7)(w^2+49)\\\\------------------------------\\\\a^2-b^2=(a-b)(a+b)$

FromTheMoon [43]3 years ago

5 0

$w^4-2.401 \\ =(w^2+49)(w^2-49) \\ =(w^2+49)(w-7)(w+7)$

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If you were to open an account that will pay you 3.5% interest with a deposit of $150 so find the following ending amount for ea

aev [14]

Answer:

FV= $160.68

Step-by-step explanation:

Giving the following information:

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To calculate the future value, we need to use the following formula:

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3 years ago

Find the least common denominator for these two rational expressions -5/s -1/2s

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3 years ago

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murzikaleks [220]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

If f(x) = 2x + 1 and g(x) = f(x) + 3, what would be the function of g(x)?

Maslowich

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2 years ago

What is the solution to log2(2x^3-8)-2log2(x)=log2(x)

Charra [1.4K]

$\bf \textit{Logarithm of rationals}
\\\\
log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y)
\\\\\\
\textit{Logarithm of exponentials}
\\\\
log_a\left( x^b \right)\implies b\cdot log_a(x)\\\\
-------------------------------$

$\bf log_2(2x^3-8)-2log_2(x)=log_2(x)
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log_2(2x^3-8)-log_2(x^2)=log_2(x)\implies log_2\left( \cfrac{2x^3-8}{x^2} \right)=log_2(x)
\\\\\\
\textit{since both sides are getting }log_2\textit{ we can simply undo that}
\\\\\\
\cfrac{2x^3-8}{x^2}=x\implies 2x^3-8=x^3\implies x^3-8=0
\\\\\\
x^3=8\implies x=\sqrt[3]{8}\implies x=2$

5 0

3 years ago

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