![\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases} 343=7\cdot 7\cdot 7\\ \qquad 7^3\\ 36=6\cdot 6\\ \qquad 6^2\\ 256=4\cdot 4\cdot 4\cdot 4\\ \qquad 4^4 \end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}} \\\\\\ \sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4} \\\\\\ 7^2+6-4^3\implies 49+6-64\implies -9](https://tex.z-dn.net/?f=%5Cbf%20343%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B36%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-256%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cbegin%7Bcases%7D%0A343%3D7%5Ccdot%207%5Ccdot%207%5C%5C%0A%5Cqquad%207%5E3%5C%5C%0A36%3D6%5Ccdot%206%5C%5C%0A%5Cqquad%206%5E2%5C%5C%0A256%3D4%5Ccdot%204%5Ccdot%204%5Ccdot%204%5C%5C%0A%5Cqquad%204%5E4%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%20%287%5E3%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B%286%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-%284%5E4%29%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B%287%5E3%29%5E2%7D%2B%5Csqrt%5B2%5D%7B%286%5E2%29%5E1%7D-%5Csqrt%5B4%5D%7B%284%5E4%29%5E3%7D%5Cimplies%20%5Csqrt%5B3%5D%7B%287%5E2%29%5E3%7D%2B%5Csqrt%5B2%5D%7B%286%5E1%29%5E2%7D-%5Csqrt%5B4%5D%7B%284%5E3%29%5E4%7D%0A%5C%5C%5C%5C%5C%5C%0A7%5E2%2B6-4%5E3%5Cimplies%2049%2B6-64%5Cimplies%20-9)
to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
Answer:
A) 0
Step-by-step explanation:
When x is divided by 11, we have a quotient of y and a remainder of 3
x/11 = y + 3
x = 11y + 3 ........(1)
When x is divided by 19, we have a remainder of 3 also
x/19 = p + 3 (p = quotient)
x = 19p + 3 ..........(2)
Equate (1) and (2)
x = 11y + 3 = 19p + 3
11y + 3 = 19p + 3
11y = 19p + 3 -3
11y = 19p
Divide both sides by 11
11y/11 = 19p/11
y = 19p/11
y and p are integers. 19 is a prime number. P/11 is also an integer
y = 19(integer)
This implies that y is a multiple of 19. When divided by 19, there is no remainder. The remainder is 0
Answer:
8.9%
Step-by-step explanation:
Here, we are to calculate the probability of Howard choosing a chocolate candy followed by a gummy candy.
The probability of selecting a chocolate candy = number if chocolate candy/ total number of candy
Total number of candy = 5 + 4 + 6 = 15
Number of chocolate candy = 5
The probability of selecting a chocolate candy = 5/15 = 1/3
The probability of selecting a gummy candy = number of gummy candies/total number of candies
Number of gummy candy = 4
The probability of selecting a gummy candy = 4/15
The probability of selecting a chocolate candy before a gummy candy = 1/3 * 4/15 = 4/45 = 0.088888888889
Which is same as 8.89 percent which is 8.9% to the nearest tenth of a percent
Answer:
2/5.
Step-by-step explanation:
The prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
A total of 8.
So the probability of rolling a prime number is 8/20
= 2/5.
