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ludmilkaskok [199]
2 years ago
10

Test |

Mathematics
1 answer:
zheka24 [161]2 years ago
6 0

Step-By-Step Explanation:

1. 20.95 x 1 = 20.95

2. 119.5 x 2 = 239

3. 8.75 x 5 = 77.5

4. 24.25 x 7 = 169.75

5. 52.3 x 3 = 156.9

6. 0.4

7. 0.4

8. 0.4

9. 0.7

10. 0.5

<em><u>If this helped, please consider picking this answer as the Brainliest Answer. Thank you!</u></em>

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BRAINLIEST ANSWER TO ANYONE WHO HELPS ME FIND QUESTIONS 1,2 AND 3
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1. a)CK²=CH²+HK²,  CH - radius
CH²=CK²-HK²=50²-30²=2500-900=1600 ft²
CH=40 ft  -- radius
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A force of 128 lb is required to hold a spring stretched 2 ft beyond its natural length. How much work W is done in stretching i
Agata [3.3K]

We have that the  workdone in stretching natural length to is mathematically given as

W=9/2lbft

<h3>Workdone in stretching natural length</h3>

Question Parameters:

  • A force of 128 lb is required to hold a spring stretched 2 ft <em>beyond </em>its natural length.
  • Stretching it from its natural length to 9 inches beyond its natural length

Generally the Hookes equation for the Force   is mathematically given as

F=Kx

Where

3.2lb=2ft*k

Thereofore

k=16

Force becomes

f=kx

f=16x

Hence

W=8[x^2]^{3/4}_{0}

W=9/2lbft

For more information on Length visit

brainly.com/question/2096451

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