
+

equals

.
First, simplify

to

and also

to

. Your problem should look like:

+

.
Second, find the least common denominator of

and

to get 9.
Third, make the denominators the same as the least common denominator (LCD). Your problem should look like:

+

.
Fourth, simplify to get the denominators the same. Your problem should look like:

+

.
Fifth, join the denominators. Your problem should look like:

.
Sixth, simplify. Your problem should look like:

, which is the answer.
Answer:
![\sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%3D%28a%5E%7B2%7D%2Bb%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Step-by-step explanation:
∵∛x = (x)^1/3
∴ ![\sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%3D%28a%5E%7B2%7D%2Bb%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
So you can replace the radicals by fractional exponents
Correct question with correct options has been attached
Answer:
37
Step-by-step explanation:
Let C represent those dressed as heroes and let B represent those wearing mask
Thus;
n(C) = 39
n(D) = 45
We are told that 21 entrants were dressed as heroes wearing a mask.
Thus; n(C n D) = 21
We are also told that there were a total of 100 entries.
Thus, n(T) = 100
Now, number of entrants neither dressed as heroes nor wearing a mask is given by;
n(T) - n(C u D)
n(C u D) = n(C) + n(D) - n(C n D)
n(C u D) = 39 + 45 - 21
n(C u D) = 63
Thus;number of entrants neither dressed as heroes nor wearing a mask is: 100 - 63 = 37
67/100 is approximately 2/3 and 515 is approximately 510 so D. 2/3*510
I’m not sure if u mean it that way but the answer is correct