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➷ Find the scale factor:
15/6 = 2.5
Multiply this by the length
2 x 2.5 = 5
x = 5 in
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Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or
ft
=
ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
=
ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k(
)
k = 18
Work done (W) to stretch the spring by
ft will be
W = 
= 
= ![18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0](https://tex.z-dn.net/?f=18%5B%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%5D%5E%7B%5Cfrac%7B11%7D%7B12%7D%7D_0)
= 9(
)²
= 7.56 lb-ft
Answer: 220
Step-by-step explanation:
176 dived by 4 is 44, 44 times 5 is 220 your welcome
Given :
A circle is inscribed in a square with a side length of 144.
So, radius of circle, r = 144/2 = 72 units.
To Find :
The probability that the point is inside the circle.
Solution :
Area of circle,

Area of square,

Now, probability is given by :

Therefore, the probability that the point is inside the circle is 0.785 .
Answer:
0.89h or 53 minutes
Step-by-step explanation:
2m/h / 2.25m
= 0.89h