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mezya [45]
3 years ago
13

What is the simplified form of this expression 7(^3sqrt2x)-3(^3sqrt16x)-3(^3sqrt8x)

Mathematics
1 answer:
trasher [3.6K]3 years ago
7 0

Answer:

<h2>7 \sqrt[3]{2x}  - 6 \sqrt[3]{2x}  - 6x</h2>

Solution,

7( \sqrt[3]{2x} ) - 3( \sqrt[3]{16x} ) - 3( \sqrt[3]{8x} ) \\  = 7 \sqrt[3]{2x}  - 3 \times ( \sqrt[3]{2 \times 2 \times 2 \times 2x}  - 3 \times  \sqrt[3]{2 \times 2 \times 2x}  \\  = 7 \sqrt[3]{2x}  - 3 \times (2 \sqrt[3]{2} x) - 3 \times 2x \\  = 7 \sqrt[3]{2x}  - 3 \times 2 \times  \sqrt[3]{2x}  - 3 \times 2x \\  = 7 \sqrt[3]{2x}  - 6 \sqrt[3]{2x}  - 6x

Hope this helps...

Good luck on your assignment...

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If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)
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We can expand the logarithm of a product as a sum of logarithms:

\log_dabc=\log_da+\log_db+\log_dc

Then using the change of base formula, we can derive the relationship

\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}

This immediately tells us that

\log_dc=\dfrac1{\log_cd}=\dfrac12

Notice that none of a,b,c,d can be equal to 1. This is because

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for any choice of y. This means we can safely do the following without worrying about division by 0.

\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}

so that

\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23

Similarly,

\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}

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So we end up with

\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}

###

Another way to do this:

\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}

\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}

\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12

Then

abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}

So we have

\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}

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Given the function:

f(x)=2^x

The function can also be written as:

y=2^x

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We have:

From the graph above, all possibe y values range from 0 to infinity.

Therefore, the range of the function is from zero infinity.

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Answer:

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Trina buys a pizza in the shape of a circle with a diameter of 14 in what is the area of the pizza​
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