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Akimi4 [234]
3 years ago
6

Tony collected data on years of employment and the annual salaries of the sales people at Company Z. He made a scatter plot and

drew a trend line that approximated the line of best fit for the data, as shown on the right. Based on the trend line drawn for the data, what is the salary a salesperson with 3 years of employment at Company Z can expect to earn?
Mathematics
1 answer:
devlian [24]3 years ago
7 0
Can you send a picture?
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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
2 years ago
Om to 4conm.
max2010maxim [7]
The correct answer is water
8 0
3 years ago
Why am this point be labeled with two names; 1 1/2 and 3/2
kakasveta [241]
3/2 is an improper fraction, where the denominator (bottom number) is smaller than the top (numerator)
1  1/2 is this same fraction but now it is not improper.  
5 0
3 years ago
Write the standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the gi
maks197457 [2]

Answer:

The formula for this quadratic function is x*2 +6x+13

Step-by-step explanation:

If we have the vertex and one point of a parabola it is possible to find the quadratic function by the use of this

y= a (x-h)*2 + K

Quadratic function looks like this

y= ax*2 + bx + c

So let's find the a

y= a (x-h)*2 + K where

y is 13, x is 0, h is -3 and K is 4

13= a (0-(-3))*2 +4

13=9a +4

9=9a

9/9=a

1=a

The quadratic function will be

y= 1(x+3)*2 + 4

Let's get the classic form

(x+3)*2 = (x+3)(x+3)

(x*2+3x+3x+9)

x*2 +6x+13

f(0) = 13

6 0
3 years ago
Read 2 more answers
The number of customers for a new online business can be modeled by y = 6x2 + 75x + 200, where x represents the number of months
qaws [65]
Plug in x=25 to get y=5825. So the answer is A.
4 0
3 years ago
Read 2 more answers
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