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elena55 [62]
3 years ago
9

How do you find the r?

Mathematics
1 answer:
astra-53 [7]3 years ago
5 0
It's a right triangle. Use the pythagorean theorem. 36^2+r^2=(r+12)^2. r=48.
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Find the dimensions of a rectangular building with a perimeter of 110 meters if the length is four
tiny-mole [99]

Answer:

b = 11 m

l = 44 m

Step-by-step explanation:

l = 4b (length is four times width)

Perimeter = 110 m

2 * ( l + b ) = 110

2 * (4b + b ) = 110

         2 * 5b = 110

                b = 110 / ( 2*5) = 11

b = 11 m

l = 4b = 4 * 11 = 44 m    

3 0
3 years ago
Naomi surveyed some students to determine their favorite subject. The experimental probability of a student choosing math as the
yawa3891 [41]
64 because if she asked some students she didn’t ask her class
8 0
3 years ago
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An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 85e−0.04t liters per minute. How much
Furkat [3]
Given the rate of water flow is given by:
r(t)=85e^(-0.04t)
thus the amount of water flow in the first hour when t=1, will be:
r(1)=85e^(-0.04*1)
=81.667
thus the amount of water that flowed out will be:
85-81.667
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5 0
3 years ago
Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
andriy [413]

Answer:

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Step-by-step explanation:

We have been given a indefinite integral \int \frac{2}{3x\left(3x-5\right)}dx. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}

\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx

\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx

Using difference rule of integrals, we will get:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)

Now, we need to use u-substitution as:

Let u=3x-5.

\frac{du}{dx}=3

dx=\frac{1}{3}du

\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|

\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|

Substitute back these values:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)

Let us add a constant C.

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Therefore, our required integral would be \frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C.

5 0
3 years ago
What is the fraction 3 is 10 into a decimal
Harlamova29_29 [7]
                    
3/10 = 0.3333333333      
6 0
3 years ago
Read 2 more answers
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