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ValentinkaMS [17]
3 years ago
8

Please help on this math question

Mathematics
1 answer:
Llana [10]3 years ago
4 0

Answer:

\large\boxed{C.\ \dfrac{-2x^2-x+72}{8x^2}}

Step-by-step explanation:

\dfrac{9}{x^2}-\dfrac{2x+1}{8x}=\dfrac{9\cdot8}{x^2\cdot8}-\dfrac{(2x+1)\cdot x}{8x\cdot x}\qquad\text{use the distributive property}\\\\=\dfrac{72}{8x^2}-\dfrac{2x^2+x}{8x^2}=\dfrac{72-(2x^2+x)}{8x^2}=\dfrac{72-2x^2-x}{8x^2}\\\\=\dfrac{-2x^2-x+72}{8x^2}

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Step-by-step explanation:

7 0
3 years ago
Help ASAP show work please thanksss!!!!
Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

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\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

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Answer:

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7 0
3 years ago
Select another format for the following Exponential Expression. 6−3
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The cube root of 6^9

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