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)
I found this online, hope it helps
Just follow these steps:
Multiply normally, ignoring the decimal points.
Then put the decimal point in the answer - it will have as many decimal places as the two original numbers combined.
Answer:
2/5÷1/6 = 2/1
Step-by-step explanation:
2/5÷1/6 = 2/5 times 6/1 = 12/5, 12÷2 is 6, 5 ÷ 5 is 1, so it's 2/1.
Answer:
yes
Step-by-step explanation:
25x² - 40xy + 16y² can be factored as
(5x - 4y)² ← a perfect square
