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sveticcg [70]
3 years ago
13

PLEASE HELP I WILL MARK YOU BRAINLIEST IF CORRECT

Mathematics
2 answers:
Maurinko [17]3 years ago
5 0

The answer you are looking for is 3x+15=5x-7. But, the answer is x=11.

Alenkinab [10]3 years ago
4 0

Sure, I will help you. The answer you are looking for is x=11.

Solution/Explanation:

Write out the equation:

3(x+5)=5x-7

Using the Distributive Property on the LEFT SIDE,

3x+15=5x-7

Next, subtracting 3x from both sides,

3x-3x+15=5x-3x-7

15=2x-7, or 2x-7=15

2x-7=15

Now, adding seven to both sides,

2x-7+7=15+7

2x=22

Finally, dividing both sides by 2,

So, therefore the final answer is x=11. And, also, the next step would be written as 3x+15=5x-7, as stated above, preliminarily.

I hope this helped you find your answer. Enjoy your day, and take care!

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[45-(6+3)]*7=<br> ???????
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The graphs of f(x) and g(x) are shown below:
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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
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