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denpristay [2]
3 years ago
7

How many minutes are there in 5/8 of 3 hours

Mathematics
2 answers:
diamong [38]3 years ago
8 0
112.5 minutes

Since there are 60 minutes in 1 hour so multiply that by 3 and you get 180 minutes
180➗8=22.5
22.5✖️5= 112.5 minutes

Hope this helps! :3
Slav-nsk [51]3 years ago
5 0
60 minutes in an hour. So 3 *60= 180

5/8*180= 5*180 = 900/8
112.5 minutes
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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
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✰ <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>

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  • 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}} \\  \\

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