Answer:
The exact answer in terms of radicals is ![x = 5*\sqrt[3]{25}](https://tex.z-dn.net/?f=x%20%3D%205%2A%5Csqrt%5B3%5D%7B25%7D)
The approximate answer is 
(accurate to 5 decimal places)
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Work Shown:
Let ![y = \sqrt[5]{x^3}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B5%5D%7Bx%5E3%7D)
So the equation reduces to -7 = 8-3y
Let's solve for y
-7 = 8-3y
8-3y = -7
-3y = -7-8 ... subtract 8 from both sides
-3y = -15
y = -15/(-3) ... divide both sides by -3
y = 5
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Since and y = 5, this means we can equate the two expressions and solve for x
![\sqrt[5]{x^3} = 5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%20%3D%205)
Raise both sides to the 5th power
![x = \sqrt[3]{3125}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B3125%7D)
Apply cube root to both sides
![x = \sqrt[3]{125*25}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B125%2A25%7D)
![x = \sqrt[3]{125}*\sqrt[3]{25}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B125%7D%2A%5Csqrt%5B3%5D%7B25%7D)
![x = \sqrt[3]{5^3}*\sqrt[3]{25}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B5%5E3%7D%2A%5Csqrt%5B3%5D%7B25%7D)
Assuming that CB is tangent, this is just a right triangle with a hypotenuse of 20 and a side of 12 so by the Pythagorean Theorem:
20^2=12^2+x^2
400=144+x^2
x^2=400-144
x^2=256
x=16
The answer is 69.6 degrees because all of the angles of a triangle always add up to 180 degrees so adding 42 and 68.4 gets 110.4, and then subtracting that from 180 gets 69.9.
Answer:
Step-by-step explanation:
imagine 38 n = lets say b
so b/b/b/b/b
and divide that by 38
A = 0.9 o
<span>so, </span>
<span>o = a/0.9 = 1.111a </span>
<span>so, the orange is 11% heavier than the apple.
glad i could help(:</span>
