What is equivalent to 3^-5

Mathematics

1 answer:

4vir4ik [10]3 years ago

4 0

Answer:

The answer is the choose: (D)

${3}^{ - 5} = \frac{1}{ {3}^{5} }$

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( I WILL BE GIVING EXTRA POINTS AND BRAINLIEST TO WHO EVER HELPS ME!! :D)

KatRina [158]

Answer:

B I think

Step-by-step explanation:

Eight eighths in one cup. Took that and multiplied that and added the single eighth. Six sixths in one cup. Took that and multiplied that and added the single sixth again. Added the total up and got 10.2916666667. Rounded. Hope this helps!

3 0

2 years ago

Read 2 more answers

Evaluate 7³=7•7•7=343

ivanzaharov [21]

343=343=343
I am pretty sure this is the answer to the problem

4 0

3 years ago

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How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$