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Ad libitum [116K]
3 years ago
9

Graph the parametric equations. x = 7 sin (t) + sin (7t) y = 7 cos (t) + cos (7t)

Mathematics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

Step-by-step explanation:

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there is no attachmengt

Step-by-step explanation:

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2 years ago
David, a high school senior was wondering if the cost of attending college is worth the
expeople1 [14]

Answer:

D.

Step-by-step explanation:

<u>Explanation for part 1.</u>

To find the mean, find total then divide by number of data set

For college salaries

Sum=41+67+53+48+45+60+59+55+52+52+50+59+44+49+52=786

Number of samples=15

Mean= 786/15 =52.4 * $1000=$52400

For High school salaries

Sum=23+33+36+29+25+43+42+38+27+25+33+41+29+33+35=492

Number of samples =15

Mean= 492/15 = 32.8 *$1000= $32800

College grads make more money according to the means.

<u>Explanation for part 2.</u>

Treat the data as part of coordinates and graph then on the same scale and axis to visualize the trend and make comparison.In this case, the graph for the line of best fit is linear as attached.

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2 years ago
Estimate the sum by first rounding each
sukhopar [10]
5 3/8 rounds to 5,
4 7/10 rounds to 5
Your estimated sum is 10
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3 years ago
Please help me quick!!
ki77a [65]

Answer: how many hours is she working

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8 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
2 years ago
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