Which is closest to the volume of this circular cone?

A. 452.4 cm3

B. 603.2 cm3

C. 150.8 cm3

D. 1,809.6 cm3

Which Is Closest To The Volume Of This Circular Cone?A. 452.4 Cm3B. 603.2 Cm3C. 150.8 Cm3D. 1,809.6 Cm3

Answers

Answer 1

Step-by-step explanation:

here's your solution

=> volume of circular cone = (1/3)πr^2h

=> radius = 4 cm ,, height = 9 cm

=> volume of cone = (1/3)*(22/7)*16*9

=> volume of circular cone = 150.8 cm^3

[tex]hope \: it \: helps[/tex]

Answer 2

Answer:

C

Step-by-step explanation:

The volume of a cone is:

(π)(r)^2(h/3)

The diameter is given as 8cm but we need the radius and the radius is half of the diameter meaning 4 cm.

And the height is 9cm

(3.14)(4)^2(9/3)

(3.14)(16)(3)

(3.14)(48)

150.72

Closest to C

Hope this helps :)


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Answers

Answer:

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Step-by-step explanation

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Answers

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Step-by-step explanation:

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Answers

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Answers

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Answers

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Answers

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4 x 3 + 2 x 9 - 40
PLEASE!!!

Answers

The answer is 86

Following PEMDAS, which is:

parentheses

exponents

multiplication

division

addition

subtraction

I found that the answer was 86

Answer:

-10.

Step-by-step explanation:

Step 1: Explain/define

This problem contains more than one operation. This means we should solve this using a special acronym known as PEMDAS.

P (Parenthesis)

E (Exponents)

M/D (Multiplication/division, whichever comes first)

A/S (Addition/subtraction, whichever comes first)

This is the correct sequence of steps when solving expressions with multiple operations.

PEMDAS can be remembered using the mnemonic "Please Excuse My Dear Aunt Sally."

------------------------------------------------------------

Step 2: Solve.

REMEMBER PEMDAS!

Expression given: 4 × 3 + 2 × 9 - 40

There are no parentheses.

There are no exponents.

There is multiplication, but no division.

4 × 3 + 2 × 9 - 40

12 + 18 - 40

There is addition and subtraction. Remember to solve whichever comes first.

12 + 18 - 40

30 - 40

30 - 40 = -10.

I, therefore, believe the answer to 4 × 3 + 2 × 9 - 40 is -10.

Please let me know if you have any further questions.

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