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maksim [4K]
2 years ago
9

Find the volume of a right circular cone that has a height of 11.9 ft and a base with a radius of 16.3 ft. Round your answer to

the nearest tenth of a cubic foot.
Mathematics
1 answer:
vivado [14]2 years ago
3 0

Answer:

3311.36ft^3

Step-by-step explanation:

Given data

Height h= 11.9ft

Radius=16.3ft

The expression for the volume of a cone is given as

Volume= 1/3πr^2h

substitute

Volume=1/3*3.142*16.3^2*11.9

Volume= 1/3*3.142*265.69*11.9

Volume= 1/3*9934.095962

Volume=3311.36ft^3

Hence the volume is 3311.36ft^3

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Choose all the expressions that are equal to 5/9×8. A. 9÷5×8 B. 8/9×5 C. 5÷8×9 D. 5×1/9×8 E. 5×8
Charra [1.4K]
<h2>Problem:</h2>

Choose all the expressions that are equal to 5/9×8.

A. 9÷5×8

B. 8/9×5

C. 5÷8×9

D. 5×1/9×8

E. 5×8

<h2>Solution:</h2>

\purple{ \frac{5}{9}  \times 8 =  \frac{40}{9}  = 4 \:  \frac{4}{9} }

\red{9 \div 5 \times 8 =  \frac{9}{5}  \times 8 =  \frac{72}{5}  = 14 \:  \frac{2}{5} }

\red{ \frac{8}{9}  \times 5 =  \frac{40}{9}  =  \boxed{4 \:  \frac{4}{9} }}

\red{5 \div 8 \times 9 =  \frac{5}{8}  \times 9 =  \frac{45}{8}  = 5 \:  \frac{5}{8} }

\red{5 \times  \frac{1}{9}  \times 8 =  \frac{5}{9}  \times 8 =  \frac{40}{9}  =  \boxed{4 \:  \frac{4}{9} }}

\red{5 \times 8 = 40}

<h2>Answer:</h2>

<u>B</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>D</u>

<h2>=============================</h2>

Hope it helps

<h2>=============================</h2>
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Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)
Tems11 [23]

The slope of the tangent line to f at x=-1 is given by the derivative of f at that point:

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