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TEA [102]
3 years ago
12

Find GH. Round to the nearest hundredth.

Mathematics
1 answer:
Bond [772]3 years ago
7 0
You need to use the cosine for this and this is how you do it.

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Find the values of x and y.
nirvana33 [79]

Answer:

x=15.75 y=59

Step-by-step explanation:

In total there is 360 degrees. We know y is 59 so we will replace it with that.

59+59=118. 360-118=242. 242/2=121. 29+29=58. So we can subtract that to equal 63. Now 63=4x. We can divide so 2x=31.5. x=15.75.

Therefore, 15.75=x and 59=y

I hope this helped ^^.

6 0
3 years ago
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Precise approximation of squar root of 12
Lady bird [3.3K]

3.46410161514. i used a calculator so hopefully this is correct

6 0
3 years ago
What is the value of<br> 1: 7³<br> 2: 2⁶<br> 3: 4⁴
Veseljchak [2.6K]
1) 343
7 • 7 • 7 = 343

2) 64
2 • 2 • 2 • 2 • 2 • 2 = 64

3) 256
4 • 4 • 4 • 4 = 256
7 0
3 years ago
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You deposit $975.52 in a bank account. Find the balance after 4 years if the account pays 7.25% annual interest compounded quart
sladkih [1.3K]

Answer:

The final ballance will be $1300.37.

Step-by-step explanation:

In this case we have a compounded interest, in order to calculate the final balance we need to use the following formula:

S = P(1 + r/n)^(n*t)

Where S is the final balance, P is the initial investment, r is the rate of interest, t is the time and n is the rate at which it is compounded. Since we have all the values we can directly apply to the formula as follows:

S = 975.52*(1 + 0.0725/4)^(4*4)

S = 975.52*(1.018125)^(16)

S = 975.52*1.333

S = 1300.37

The final ballance will be $1300.37.

4 0
3 years ago
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
4 0
3 years ago
Read 2 more answers
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