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3 years ago

The radius of a cylindrical water tank is 6.5 ft, and its height is 19 ft. What is the volume of the tank?

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

The volume of the cylinder is 2520.64ft³

Step-by-step explanation:

To calculate the volume of a cylinder we have to use the following formula:

v = volume

h = height = 19ft

π = 3.14

r = radius = 6.5ft

v = (π * r²) * h

we replace with the known values

v = (3.14 * (6.5ft )²) * 19ft

v = (3.14 * 42.25ft²) * 19ft

v = 132.665ft² * 19ft

v = 2520.64ft³

The volume of the cylinder is 2520.64ft³

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Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

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dolphi86 [110]

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What is the minimum value for g(x)=x2−10x 16? enter your answer in the box.

motikmotik

The minimum value for g(x)=x² - 10x + 16 is -9

<h3>How to determine the minimum value?</h3>

The function is given as:

g(x)=x² - 10x + 16

Differentiate the function

g'(x) = 2x - 10

Set the function 0

2x - 10 = 0

Add 10 to both sides

2x = 10

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Substitute 5 for x in g(x)

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Evaluate

g(5) = -9

Hence, the minimum value for g(x)=x² - 10x + 16 is -9

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GrogVix [38]

The ? is 1.

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