Answer:
The correct answer is if you want no roots in the denominator, the correct answer is 
* ![\sqrt[3]{2y^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2y%5E%7B2%7D%20%7D)
Step-by-step explanation:
The mistake is that you can't take a cube root of a square at step 3 when 
is taken out.
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5 : 20
Divide both sides by 5 to get the answer
5/5 = 1
20/5 = 4
Thus, the answer is 1 : 4
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Answer: Hmm Ok i do this-
Step-by-step explanation:
Wait are we trying to find x?
Ok so for the first one x = 5 and for the second one x = 2
<span>F(x) = (1/4)^x
</span><span>F(3) = (1/4)^3 = 1/64
</span><span>B. 1/64</span>
The sum of any geometric sequence, (technically any finite set is a sequence, series are infinite) can be expressed as:
s(n)=a(1-r^n)/(1-r), a=initial term, r=common ratio, n=term number
Here you are given a=10 and r=1/5 so your equation is:
s(n)=10(1-(1/5)^n)/(1-1/5) let's simplify this a bit:
s(n)=10(1-(1/5)^n)/(4/5)
s(n)=12.5(1-(1/5)^n) so the sum of the first 5 terms is:
s(5)=12.5(1-(1/5)^5)
s(5)=12.496
as an improper fraction:
(125/10)(3124/3125)
390500/31240
1775/142
