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Nonamiya [84]
3 years ago
10

A designer wants to make use of a half-spherical cistern in a water system. If the cistern has a diameter of 5.5 feet, how many

gallons of water can it hold?
Mathematics
1 answer:
mamaluj [8]3 years ago
8 0
\textit{volume of a sphere}=\cfrac{4}{3}\pi r^3\qquad r=radius
\\ \quad \\
\textit{so, a half of that, or a half-sphere volume}=\cfrac{\frac{4}{3}\pi r^3}{2}
\\ \quad \\
\cfrac{4}{3}\pi r^3\cdot \cfrac{1}{2}\implies \cfrac{4\pi r^3}{6}\implies \cfrac{2\pi r^3}{3}

plug in your radius
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Goofy wants to buy white rice. The store has 4 different sizes of white rice bags: $2.00 for 1 pound, $4.25 for 2 pounds, $9.50
Tanzania [10]

Answer:

20 dollars for 12 pounds.

Step-by-step explanation:

This is a critically important question to know the answer to. It is used almost every day by good shoppers.

The trick is to get everything down to 1 pound.

$2.00

$2.00 per pound. The question is the answer.

$4.25

$4.25 / 2 pounds = $2.13 rounded.

$9.50

$9.50 / 5 = $1.90

$20

$20/12 = $1.67 rounded

8 0
3 years ago
I need help and it’s due today
olga nikolaevna [1]

Answer: same bruh

Step-by-step explanation:

7 0
3 years ago
Identify the inverse of the following.
posledela

Answer:

f^{-1}(x)=\sqrt{\frac{1}{8}(-x+4)}

Step-by-step explanation:

Given function is,

f(x) = -8x² + 4

Step to find the inverse of the function,

Step 1

Write the function in the form of an equation,

y = -8x² + 4

Step 2

Interchange the variables 'x' and 'y',

x = -8y² + 4

Step 3

Solve the equation for y,

x = -8y² + 4

x - 4 = -8y²

-x + 4 = 8y²

y² = \frac{1}{8}(-x+4)

y=\sqrt{\frac{1}{8}(-x+4)}

Therefore, inverse of the function f(x) will be,

f^{-1}(x)=\sqrt{\frac{1}{8}(-x+4)}

3 0
3 years ago
the value of x + y is a negative, and x is a positive integer. What must be ture about y ? Explain....
dimulka [17.4K]
Y has to be less than negative X
8 0
3 years ago
What is the solution to the equation 1/x = x+3/2x^2
Dimas [21]
Hello,

As you write 1/x, x must be different of 0.
\dfrac{1}{x} = \dfrac{x+3}{2x^2} \\

==\textgreater\ 2x^2=x^2+3x \cross`\product \\

==\textgreater\ x^2-3x =0\\

==\textgreater\ x(x-3)=0 \\

==\textgreater\ x=0 (to exclude) or x=3\\

Answer\ x=3



8 0
3 years ago
Read 2 more answers
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