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Natasha_Volkova [10]
3 years ago
12

PLS HELP ILL GIVE BRAINLY

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
8 0

Answer:

Zeros to the right of the decimal place that are NOT merely place holders are significant; these significant zeros will be to the right of non-zero significant digits. The zero to the left of the decimal place in 20.00cm is also significant since it now falls between significant digits.

Hope this helps, have a nice day/night! :D

Aneli [31]3 years ago
5 0

Answer:

thanks for points......

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Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

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\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
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Answer:

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Answer:

Hello!

After reviewing the question you have provided, I came up with the correct answer:

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