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Dmitry [639]
3 years ago
15

The longest spin of a basketball on one finger is 255 minutes.about how many hours is that

Mathematics
1 answer:
lidiya [134]3 years ago
4 0

The longest spin of a basketball on one finger is 255 minutes or 4.25 hours.

Step-by-step explanation:

the longest spin of a basketball on one finger is 255 minutes.

We need to find how many hours is that.

So, we need to convert 255 minutes into hours.

We know that,

60 minutes = 1 hour

Using Unitary method:

60 minutes = 1 hour

1 minute = 1/60 hour

255 minutes = 1/60 * 255 hour

255 minutes = 255/60 hour = 4.25 hour

255 minutes is 4.25 hour.

So, The longest spin of a basketball on one finger is 255 minutes or 4.25 hours.

Keywords: Convert Minutes to Hour

Learn more about Convert Minutes to Hour at:

  • brainly.com/question/2821386
  • brainly.com/question/3398261
  • brainly.com/question/5248528

#learnwithBrainly

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*Hope this helps .^.*

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I don't know if this is right... please someone help mee
worty [1.4K]
For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
6 0
3 years ago
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