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castortr0y [4]
3 years ago
8

If the height and radius of a cylinder are doubled, how will the volume of the cylinder be affected?

Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0

Hope this helps!!!

Have a great day!!!

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Pythagorean Theorem is...   C) true for all right triangles

Step-by-step explanation:

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Is -33 a whole number?​
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The indoor climbing gym is a popular field trip destination. This year the senior class at High School A and the senior class at
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A wire that is 22 feet long connects the top of a pole to the ground. The wire is attached to the ground at a point that is 10 f
krok68 [10]

✰ <u>Concept</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>

⠀

In this question, we can clearly observer that the diagram shows a right angled triangle. And, we have been provided with the value of base, and the value of hypotenuse, using the pythagoras theorem, now we can easily find out the value of the perpendicular i.e. the value of the side h. According to the pythagoras theorem, square of hypotenuse is equal to the sum of square of perpendicular and square of side respectively. Therefore, square of side is equal to the difference of square of hypotenuse and square of perpendicular.

⠀

✰ <u>Given</u><u> </u><u>Information</u><u> </u>:-

⠀

  • Hypotenuse = 22 ft.
  • Base = 10 ft.

⠀

✰ <u>To Find</u><u> </u><u>:</u><u>-</u>

⠀

  • The value of side or the perpendicular

⠀

✰ <u>Formula</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>

⠀

\star \:  \underline{ \boxed{ \purple { \sf  {Side}^{2}  =  {Hypotenuse}^{2}  -  {Base}^{2}  }}} \:  \star

⠀

✰ <u>Solution</u><u> </u><u>:</u><u>-</u>

⠀

\sf \longrightarrow  {Side}^{2} =   {(22 \: ft)}^{2}  -  {(10 \: ft)}^{2}  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{2} =   {484 \: ft}^{2}  -  {100 \: ft}^{2}  \:  \:  \: \:  \:  \:   \\  \\  \\ \sf \longrightarrow  {Side}^{2} =   {384 \: ft}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{} =   \sqrt{ {384 \: ft}^{2} }  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\ \sf \longrightarrow  {Side}^{} =   \underline{ \boxed{ \frak{ \green{19.60 \: ft}}}} \:  \star \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\

Thus, option B. 19.60 ft. is the correct option.

⠀

\underline{\rule{230pt}{2pt}} \\  \\

4 0
3 years ago
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