Answer:
4.5 ≤x<4.55
Step-by-step explanation:
for whole number
It has to be 4.5 or greater to round to 5 on the low end
It has to be less than 5.5 to round to 5 on the high end
for tenths
If has to be 4.45 or greater to round to 4.5 on the low end
It has to be less than 4.55 to round to 4.5
We take the more restrictive limits to define the number
The number must be between
4.5 ≤x<4.55
6x + x + x - 5 - 2 = 8 + 2x + x
combine like terms
8x -7 = 8 +3x
subtract 3x from each side
5x -7 = 8
add 7 to each side
5x = 15
divide by 5
x=3
Answer: 8x -7 = 8 +3x
x=3
![\left(\dfrac{3a^{-3}b^2}{2a^{-1}b^0}\right)^2](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B3a%5E%7B-3%7Db%5E2%7D%7B2a%5E%7B-1%7Db%5E0%7D%5Cright%29%5E2)
First some simplifying:
![\left(\dfrac{3a^1b^2}{2a^3b^0}\right)^2=\left(\dfrac{3b^2}{2a^2}\right)^2=\dfrac{9b^4}{4a^4}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B3a%5E1b%5E2%7D%7B2a%5E3b%5E0%7D%5Cright%29%5E2%3D%5Cleft%28%5Cdfrac%7B3b%5E2%7D%7B2a%5E2%7D%5Cright%29%5E2%3D%5Cdfrac%7B9b%5E4%7D%7B4a%5E4%7D)
Then with
and
, we have
and
, so
![\dfrac{9b^4}{4a^4}=\dfrac{9\cdot9^2}{4\cdot4^2}=\dfrac{9^3}{4^3}=\dfrac{729}{64}](https://tex.z-dn.net/?f=%5Cdfrac%7B9b%5E4%7D%7B4a%5E4%7D%3D%5Cdfrac%7B9%5Ccdot9%5E2%7D%7B4%5Ccdot4%5E2%7D%3D%5Cdfrac%7B9%5E3%7D%7B4%5E3%7D%3D%5Cdfrac%7B729%7D%7B64%7D)