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topjm [15]
3 years ago
5

Looking for the answer for problem number eight zoom in if needed

Mathematics
2 answers:
Law Incorporation [45]3 years ago
5 0

For this, we will be using Triangle Angle Sum Theorem (all interior angles in a triangle add up to 180°) for Triangle BCD. Since Angle CBD and BDC are congruent to each other and Angle BDA, we can solve for those two angles to get that angle. Our equation will look like this: 180=35+2x

Firstly, subtract 35 on both sides of the equation: 145=2x

Next, divide both sides by 2 and your answer will be 72.5 = x.

Since Angle CBD and BDC are 72.5°, this means that Angle BDA is 72.5° as well.

Deffense [45]3 years ago
5 0
If < BCD IS 35 degrees
< ECD is exactly half of that 35/2= 17.5

< CED angle is 90 degrees

Because we know that in a triangle all angles combined add to 180 degrees that means that :

< EDC = 180 - 90 - 17,5 = 72,5 degrees

We know that the angle of < BDA is a mirrored angle of
m
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Michelle is painting wooden blocks. he has 11/12 of a quart of paint each block needs 1/6 of a quart to paint. How many blocks c
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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
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

to

\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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2 years ago
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