Equation would be y=x+5/8
we have
![x^{2}+2x + 1 > 0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B2x%20%2B%201%20%3E%200)
we know that
![x^{2}+2x+1=(x+1)^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B2x%2B1%3D%28x%2B1%29%5E%7B2%7D)
so
![(x+1)^{2}> 0](https://tex.z-dn.net/?f=%28x%2B1%29%5E%7B2%7D%3E%200)
two solutions
<u>first solution </u>
![(x+1)> 0](https://tex.z-dn.net/?f=%28x%2B1%29%3E%200)
![x > -1](https://tex.z-dn.net/?f=x%20%3E%20-1)
the first solution is the interval---------> (-1,∞)
<u>second solution</u>
![-(x+1)> 0](https://tex.z-dn.net/?f=-%28x%2B1%29%3E%200)
![-x-1> 0](https://tex.z-dn.net/?f=-x-1%3E%200)
![-x> 1](https://tex.z-dn.net/?f=-x%3E%201)
![x< -1](https://tex.z-dn.net/?f=x%3C%20-1)
the second solution is the interval---------> (-∞,-1)
therefore
the solution is all real numbers except the number ![-1](https://tex.z-dn.net/?f=-1)
the answer in the attached figure
Answer:
Step-by-step explanation:
nth term = ar^(n-1)
a3 = 1/8; ar² = 1/8
a6 = 1; ar⁵ = 1
![ar^{2}*r^{3}=1\\\\\frac{1}{8}*r^{3}=1\\\\r^{3}=1*8\\\\r^{3}=8\\\\r^{3}=2^{3}\\\\r=2](https://tex.z-dn.net/?f=ar%5E%7B2%7D%2Ar%5E%7B3%7D%3D1%5C%5C%5C%5C%5Cfrac%7B1%7D%7B8%7D%2Ar%5E%7B3%7D%3D1%5C%5C%5C%5Cr%5E%7B3%7D%3D1%2A8%5C%5C%5C%5Cr%5E%7B3%7D%3D8%5C%5C%5C%5Cr%5E%7B3%7D%3D2%5E%7B3%7D%5C%5C%5C%5Cr%3D2)
![ar^{2}=\frac{1}{8}\\\\a*2^{2}=\frac{1}{8}\\\\a*4=\frac{1}{8}\\\\a=\frac{1}{8*4}\\\\a=\frac{1}{32}\\\\](https://tex.z-dn.net/?f=ar%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5Ca%2A2%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5Ca%2A4%3D%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5Ca%3D%5Cfrac%7B1%7D%7B8%2A4%7D%5C%5C%5C%5Ca%3D%5Cfrac%7B1%7D%7B32%7D%5C%5C%5C%5C)
Given:
The expression is
![2b+5+1b=3b+5](https://tex.z-dn.net/?f=2b%2B5%2B1b%3D3b%2B5)
To find:
Whether the two expression in the equation are equal or equivalent.
Solution:
If two expression are exactly the same, then they are equal and if the two expressions are different but the after simplification both are same, then they are called equivalent expressions.
We have,
![2b+5+1b=3b+5](https://tex.z-dn.net/?f=2b%2B5%2B1b%3D3b%2B5)
Taking LHS, we get
![LHS=2b+5+1b](https://tex.z-dn.net/?f=LHS%3D2b%2B5%2B1b)
On combining liker terms, we get
![LHS=(2b+1b)+5](https://tex.z-dn.net/?f=LHS%3D%282b%2B1b%29%2B5)
![LHS=3b+5](https://tex.z-dn.net/?f=LHS%3D3b%2B5)
![LHS=RHS](https://tex.z-dn.net/?f=LHS%3DRHS)
In the given equation both expression are different but after simplification LHS = RHS, therefore the expression are equivalent not equal.
V = 4/3 (3.14) (5) to the third power 3
(5)3rd = 125
4/3 x (3.14 x 125)
3.14 x 125 = 392.5
4/3 x 392.5 = 1570 / 3
Volume = 523.33