Answer:
The simplified version of
is
.
Step-by-step explanation:
The given expression is
![\sqrt[3]{135}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D)
According to the property of radical expression.
![\sqrt[n]{x}=(x)^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%3D%28x%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Using this property we get
![\sqrt[3]{135}=(135)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%28135%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\sqrt[3]{135}=(27\times 5)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%2827%5Ctimes%205%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\sqrt[3]{135}=(3^3\times 5)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%283%5E3%5Ctimes%205%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![[\because (ab)^x=a^xb^x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28ab%29%5Ex%3Da%5Exb%5Ex%5D)
![[\because \sqrt[n]{x}=(x)^{\frac{1}{n}}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5Bn%5D%7Bx%7D%3D%28x%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5D)
![\sqrt[3]{135}=3\sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D3%5Csqrt%5B3%5D%7B5%7D)
Therefore the simplified version of
is
.
30,77,55,71,19,40,10,5,7,6
Answer:
a(n)=1.15[a(n-1)]
Step-by-step explanation:
we know that

Let
a0 -----> the length of the original copy
<em>The first copy is equal to</em>
a1=1.15(a0)
<em>The second copy is </em>
a2=1.15[1.15(a0)] or a2=1.15[a1]
<em>The third copy is</em>
a3=1.15{1.15[1.15(a0)]} or a3=1.15[a2]
therefore
A recursive formula will be
a(n)=1.15[a(n-1)]
2304π is the answer and if you don’t want pi it’s 7238.2294738709 feet
Answer:
46 terms
Step-by-step explanation:
a₁=b, a₂₁= bₙ, d₁=9, d₂=4
n=?
---------------
a₂₁= a₁+20d₁= a₁+20*9= a₁+180
bₙ= b₁+(n-1)d₂= a₁+4(n-1)
a₁+4(n-1)=a₁+180
4(n-1)=180
n-1= 180/4
n-1= 45
n=46