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S_A_V [24]
3 years ago
9

Simplify the following:

Mathematics
2 answers:
g100num [7]3 years ago
8 0

Answer:

\sqrt{ \frac{15x}{ {x}^{3} } }= \sqrt{ \frac{15}{ {x}^{2} } }  =   \frac{ \sqrt{15} }{ \sqrt{ {x}^{2} } }  =  \frac{ \sqrt{15} }{x}  \\

<h3>(√15)/x is the right answer.</h3>

\frac{ \sqrt{ {x}^{2} } }{\sqrt{{x}^{3}}}  =  \frac{ {x}^{ \frac{2}{2} } }{ {x}^{ \frac{3}{2}}}=  \frac{x}{ {x}^{(1 +  \frac{1}{2}) } }  =  \frac{x}{x \times  {x}^{ \frac{1}{2} } }  =  \frac{1}{ \sqrt{x} }  \:  \\

<h3>1/(√x) is the right answer.</h3>

Ket [755]3 years ago
5 0

Answer:

For the first question:

\sqrt{\frac{15x}{x^3}}\\= \sqrt{\frac{15}{x^2}}\\= \frac{\sqrt{15}}{x}\\

And the second

\frac{\sqrt{x^2}}{\sqrt{x^3}}\\= (x^2)^{\frac{1}{2} }(x^3)^{\frac{-1}{2} }\\= x^1 \times x^{\frac{-3}{2} }\\= x^{\frac{2}{2}} \times x^{\frac{-3}{2} }\\= x^{-\frac{1}{2}}\\= \frac{1}{\sqrt{x}}

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viktelen [127]

Answer:

<u>Domain:</u>

The domain of this can be any value between 0 to 565 miles per hour

<u>Range:</u>

The reasonable range can be the distance traveled which can be from 0 to 13,560 miles (no plane travel is longer than 24 hours, we assume).

Step-by-step explanation:

Domain is the input, set of x values for the function.

Range is the output, set of y values for the function.

This isn't a function essentially, but it is given that an Airplane travels at 565 miles per hour.

<em>We can say that the domain will be the speed of the airplane and the range would be the distance it travels.</em>

<em />

<u>Domain:</u>

The domain of this can be any value between 0 to 565 miles per hour

<u>Range:</u>

The reasonable range can be the distance traveled which can be from 0 to (565*24=13,560 miles) 13,560 miles (no plane travel is longer than 24 hours, we assume).

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luda_lava [24]
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