Answer:
![2a^3b^2\sqrt[3]{3a}](https://tex.z-dn.net/?f=2a%5E3b%5E2%5Csqrt%5B3%5D%7B3a%7D)
Step-by-step explanation:
Use the following rules for exponents:
![a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x](https://tex.z-dn.net/?f=a%5Em%2Aa%5En%3Da%5E%7Bm%2Bn%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7Bx%5E3%7D%3Dx)
Simplify 24. Find two factors of 24, one of which should be a perfect cube:
Insert:
![\sqrt[3]{2^3*3a^{10}b^6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%5E3%2A3a%5E%7B10%7Db%5E6%7D)
Now split the exponents. Split 10 into as many 3's as possible:
Insert as exponents:
![\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%5E3%2A3%2Aa%5E3%2Aa%5E3%2Aa%5E3%2Aa%5E1%2Ab%5E6%7D)
Split 6 into as many 3's as possible:
Insert as exponents:
![\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%5E3%2A3%2Aa%5E3%2Aa%5E3%2Aa%5E3%2Aa%5E1%2Ab%5E3%2Ab%5E3%7D)
Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):
![2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}](https://tex.z-dn.net/?f=2%5Csqrt%5B3%5D%7B3%2Aa%5E3%2Aa%5E3%2Aa%5E3%2Aa%5E1%2Ab%5E3%2Ab%5E3%7D%5C%5C%5C%5C%5C%5C2%2Aa%2Aa%2Aa%5Csqrt%5B3%5D%7B3%2Aa%5E1%2Ab%5E3%2Ab%5E3%7D%5C%5C%5C%5C%5C%5C2%2Aa%2Aa%2Aa%2Ab%2Ab%5Csqrt%5B3%5D%7B3%2Aa%5E1%7D)
Simplify:
:Done
7 is infinite 6 is 4/3 5 is zero 4 is negative 5/4
Answer:
x=60
Step-by-step explanation:
Hope this helped.
Assuming that the triangles are congruent, you can just look at the parts on each triangle to find the parts on the other triangle. The answers are yes, yes, and no. The triangles are obviously not drawn to scale, so just remember to never rely on it being so.
You would need 27 more dollars because f(x)=7x+2 plug in the 5 in x’s spot and you get 37 and so subtract it from 10 to get 27
