Answer:
Your correct and final answer is (9+x^6) (3+x^3) (3-x^3)
Hope this helps!!!
Answer:
The area covered by the planters is 113.6 square inches.
Step-by-step explanation:
Assuming that the dimmensions of the greenhouse are 4 1/4 by 1 2/3 inches. We first need to convert them from mixed fractions to normal fractions, we do that by summing the integer part to the fraction one. We have:
length = 4 + 1/4 = 16/4 + 1/4 = 17/4 inches
width = 1 + 2/3 = 3/3 + 2/3 = 5/3 inches
The area of the greenhouse is given by:
area = length*width
area = (17/4)*(5/3) = 85/12 = 7.1 square inches
Since there is an array of 4x4 there is a total of 16 greenhouses and it's total area is 16*7.1 = 113.6 square inches
Answer: The second option (purple model)
Step-by-step explanation: It is that option because there were 80 toys collected which is 100% and then we do not what the 30% is.
Answer:
Option d) 5 to the power of negative 5 over 6 is correct.
![\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B%5Cbf%205%7D%20%5Ctimes%20%5Csqrt%7B%5Cbf%205%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cbf%205%5E%7B%5Cbf%205%7D%7D%7D%3D%205%5E%7B%5Cfrac%7B%5Cbf%20-5%7D%7B%5Cbf%206%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
ie, 
Step-by-step explanation:
Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
It can be written as below
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B2%2B3%7D%7B6%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5%7D%7B6%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B-5%7D%7B3%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5-10%7D%7B6%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B5%5E5%7D%3D%205%5E%7B%5Cfrac%7B-5%7D%7B6%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
The answer is A.
I think so because each one of them agrees with 20x and 15
