Answer:
<h2>Hi there !</h2>

![y = \sqrt[2]{x}+2 \: will \: pass \: through \: (0 \:, 2) \: and \: ( - 8 \:, 0)](https://tex.z-dn.net/?f=y%20%3D%20%20%5Csqrt%5B2%5D%7Bx%7D%2B2%20%5C%3A%20will%20%5C%3A%20pass%20%5C%3A%20through%20%5C%3A%20%280%20%5C%3A%2C%202%29%20%5C%3A%20and%20%5C%3A%20%28%20-%208%20%5C%3A%2C%200%29)
Step-by-step explanation:
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Answer:
Change in area=24
-48
Step-by-step explanation:
Let s will be the side of square and r will be the radius of circle.
Then two given conditions are
1)dr/dt=2 m/s
2)ds/dt=1 m/s
Area enclosed=(Area of square)-(Area of circle)
Area of square=
Area of circle=
Area enclosed=
dA/dt=2
r(dr/dt)-2s(ds/dt)
At s=24,and r=6
dA/dt=2(
)(6)(2)-2(24)(1)
Change in area=24
-48
1 m = 100 cm, so 5 m = 500 cm plus the 32 cm, so she has a rope that is 532 cm long.
532 - 249 = 283 cm
So the other piece of rope is 283 cm long.