Answer:
29
Step-by-step explanation:
Total marble = 36
Let red marbles be X
Then green marble is 4X + 1
And 4X + 1 + X = 36
5X = 36 - 1
5X = 35
X = 35/5 = 7
So, green marbles = 4X + 1
= 4(7) + 1
= 29
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:
well all of these look like a way so we have to use elimination method
A : random number tables : well it has random numbers so X out
B: PHONE NUMBERS: well phone numbers are random so X out
C: USing the internet : totally X out
D: books of random numbers: X out
so none of the above i guess
Do the division
(5.685 * 10^26) / (3.302*10^23)
= 1.722 * 10^3
or 1722 times greater
