Answer:
angle NQT equals angle STR
Step-by-step explanation:
You already have two angLes, you just need a side to be able to say the AAS Theorem. The answer is the remaining side to figure that out
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
Supplementary = 180°
Complementary = 90°
180° - 52° = 128°
90° - 52° = 38°
Hope this Helps!!
Let x + 1.5 be length of segment above left hand
Let 2x be length of segment below left hand
15 = x + 1.5 + 2x
15 = 3x + 1.5
15 - 1.5 = 3x
13.5 = 3x
13.5/3= 3x/3
4.5 = x
x = 4.5
This is the length above the left hand.
2x must be then 4.5*2 = 9 foot
This is the length below the left hand.
Checking
4.5 + 9 + 1.5 = 15 foot
Adding length below left hand & length of left hand to right hand:
9 + 1.5 = 10.5
The right hand is hence 10.5 foot far up the pole.
