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3 years ago

Help me, save my life.

Mathematics

1 answer:

NemiM [27]3 years ago

4 0

25^2 or 5 hope this helps

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A handyman company serves customers from two cities 60 miles apart, Jonesville and Bellevue. The company would like to position

maw [93]

The new office should be at a distance of 37.5 miles from Jonesville.

The weighted mean is the sum of all elements in a data-set multiplied by their weights, divided by the sum of the weights. For example, if x1 and x2 are the elements with weights w1 and w2 respectively, then the weighted average is calculated using the given formula-

M = {(x1*w1) + (x2*w2)}/ (w1+w2)

Here, we are given that two cities Jonesville and Bellevue, 60 miles apart.

Let Jonesville be at point 0 and Bellevue be at point 60.

Jonesville has a weight of 3 and Bellevue has a weight of 5.

Thus, the weighted average can be calculated as follows-

Mean = {(60*5) + (0*3)}/8

= 300/8

= 75/2

= 37.5

Thus, the new office should be at a distance of 37.5 miles from Jonesville.

Learn more about weighted averages here-

brainly.com/question/28507971

#SPJ1

4 0

1 year ago

Please help me with the below question.

Alexus [3.1K]

a) Substitute $y=x^9$ and $dy=9x^8\,dx$ :

$\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}$

b) Integrate by parts:

$\displaystyle \int u\,dv = uv - \int v \, du$

Take $u = \ln(x)$ and $dv=\frac{dx}{x^7}$ , so that $du=\frac{dx}x$ and $v=-\frac1{6x^6}$ :

$\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}$

c) Substitute $y=\sqrt{x+1}$ , so that $x = y^2-1$ and $dx=2y\,dy$ :

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy$

Integrate by parts with $u=y$ and $dv=e^y\,dy$ , so $du=dy$ and $v=e^y$ :

$\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C$

Then

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}$

d) Integrate by parts with $u=\sin(\pi x)$ and $dv=e^x\,dx$ , so $du=\pi\cos(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx$

By the fundamental theorem of calculus,

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx$

Integrate by parts again, this time with $u=\cos(\pi x)$ and $dv=e^x\,dx$ , so $du=-\pi\sin(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx$

By the FTC,

$\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx$

Then

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}$

e) Expand the integrand as

$\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}$

Then by the FTC,

$\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}$

f) Substitute $e^{7x} = \tan(y)$ , so $7e^{7x} \, dx = \sec^2(y) \, dy$ :

$\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}$

8 0

2 years ago

Tom's father gave Tom $100 for lunch at school. After 10 days,

galina1969 [7]

Answer:

He spent 10 dollars

Step-by-step explanation:

Because, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.

And that's 10 days, so he spent 10 dollars for everyday. hope that helps!

Sorry if its wrong.

5 0

3 years ago

Is - (- 1 1/2 ) located to the right or left of zero?

Bess [88]

As the number -1 1/2 would be to the left because it is negative, if it is as the number's integer, then the new number would be 1 1/2 because 1 1/2 and -1 1/2 are integers.

Hope this helps, if not, comment below please!!!!

5 0

3 years ago

How do you find the radius of a circle

tia_tia [17]

Answer:

To calculate the radius of a circle by using the circumference, take the circumference of the circle and divide it by 2 times π.

3 0

4 years ago

Read 2 more answers