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erastova [34]
3 years ago
13

The area of a right triangle is 60 ft2. The base of the triangle is 8 ft. What is the length of the hypotenuse? Show your work.

Don’t forget to label your answer.
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
4 0
I think the hypotenuse is 17
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yaroslaw [1]

Answer:

9pi

Step-by-step explanation:

circumference= 2 (pi) (r)

radius = diameter \ 2

r = 9 \ 2 = 4.5in.

c = 2 (pi) (4.5)

c = 9pi or 28.3in.

Hope this helps!

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Write 2/3 of 4 as a fraction
eduard
 2 / 3 of 4 = ( 2 x 4 ) / 3 = 8 / 3 ;
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A survey on planet rew found that 80% preferred tott juice to all other juices. if 32 beings preferred tott juice, how many bein
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2 years ago
Can someone help me please
kifflom [539]

9514 1404 393

Answer:

  {Segments, Geometric mean}

  {PS and QS, RS}

  {PS and PQ, PR}

  {PQ and QS, QR}

Step-by-step explanation:

The three geometric mean relationships are derived from the similarity of the triangles the similarity proportions can be written 3 ways, each giving rise to one of the geometric mean relations.

  short leg : long leg = SP/RS = RS/SQ   ⇒   RS² = SP·SQ

  short leg : hypotenuse = RP/PQ = PS/RP   ⇒   RP² = PS·PQ

  long leg : hypotenuse = RQ/QP = QS/RQ   ⇒   RQ² = QS·QP

I find it easier to remember when I think of it as <em>the segment from R is equal to the geometric mean of the two segments the other end is connected to</em>.

__

  segments PS and QS, gm RS

  segments PS and PQ, gm PR

  segments PQ and QS, gm QR

4 0
3 years ago
What is the sum of the infinite geometric series?<br><br> 120 + 20+ 10/3 + 5/9+...
Sergeu [11.5K]

Answer:

for an infinite geometric series the formula for the sum of the infinite geometric series when the common ratio is less than one is given by

\frac{a_{1} }{1 - r} = \frac{120}{1 - \frac{1}{6} } = \frac{120}{\frac{5}{6} } = \frac{120 \times 6}{5} = 144.

Step-by-step explanation:

i) from the given series we can see that the first term is a_{1 } = 120.

ii) let the common ratio be r.

iii) the second term is 20 = 120 × r

   therefore r = 20 ÷ 120 = \dfrac{1}{6}

iv) the third term is \frac{10}{3} = 20 × r

    therefore r = \dfrac{10}{3} ÷ 20 = \dfrac{1}{6}

v) for an infinite geometric series the formula for the sum of the infinite geometric series when the common ratio is less than one is given by

\frac{a_{1} }{1 - r} = \frac{120}{1 - \frac{1}{6} } = \frac{120}{\frac{5}{6} } = \frac{120 \times 6}{5} = 144.

6 0
3 years ago
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