When you have angle and its corresponding (opposite) side then you need to use law of sin. Problems 3,4,5(in 5 you need at first find angle X=180-92-62)
when you have two sides and angle between them, this is law of cos, problem 6,8
problem 7 you can use both, but if you can use both better to use law of sin, it is just easier
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
Answer:
9.9x + 16.1
Step-by-step explanation:
SImplify using like terms
Simplify:
- 3.2x + 7.9 + 8.2 + 6.7x
- 3.2x + 6.7x + 7.9 + 8.2
- 9.9x + 16.1
-Chetan K
Yes it is. A gallon is 4 quarts, so he has more than enough. He has 2 quarts extra.
Answer:
A right angle.
Step-by-step explanation:
65+25 =90
