Remember that you cannot take the log of a negative number, regardless of the base. Also, you can take the logarithm of zero, since no number raised to a power can equal zero (although it come very close).
Therefore, (x+3), the number inside the logarithm, must be greater than zero.
x+3 > 0
x > -3
The domain is x > -3,
or (-3, ∞) in interval notation
The answer should be SAS, since you know two sides and an angle in the boy’s case, and a side and an angle in the tree’s case.
If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785_______________
Evaluate the indefinite integral:

Make a trigonometric substitution:

so the integral (i) becomes


Now, substitute back for t = arcsin(x²), and you finally get the result:

✔
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You could also make
x² = cos t
and you would get this expression for the integral:

✔
which is fine, because those two functions have the same derivative, as the difference between them is a constant:
![\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\ =\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29-%5Cleft%28-%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%5Cright%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29%2B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%5B%5C%2Carcsin%28x%5E2%29%2Barccos%28x%5E2%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7D)

✔
and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.
I hope this helps. =)
6 and 10 are 2 units away from a positive 8 on the number line
Answer:
1st 306.36 2nd 10.78 3rd 355.25
Step-by-step explanation:
bc i just skipped and it showed me the answers