Answer:

Step-by-step explanation:
This question is on rules of rational exponential
where the exponential is a fraction, you can re-write it using radicals where the denominator of the fraction becomes the index of the radical;
<u>General expression </u>
![a^\frac{1}{n} =\sqrt[n]{a}](https://tex.z-dn.net/?f=a%5E%5Cfrac%7B1%7D%7Bn%7D%20%3D%5Csqrt%5Bn%5D%7Ba%7D)
Thus ![\sqrt[3]{x} =x^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%20%3Dx%5E%5Cfrac%7B1%7D%7B3%7D)
<u>Applying the same in the question </u>
![\sqrt[3]{x^5y} =x^\frac{5}{3} y^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E5y%7D%20%3Dx%5E%5Cfrac%7B5%7D%7B3%7D%20y%5E%5Cfrac%7B1%7D%7B3%7D)
=
Answer:
3/10 of the aquarium are yellow tangs.
Step-by-step explanation:
Given that in an aquarium 2/5 of the fish or surgeonfish and of these, 3/4 are yellow tangs, to determine what fraction of all fish in the aquarium are yellow tangs the following calculation must be performed:
2/5 = 0.40
0.40 x 3/4 = X
1.20 / 4 = X
0.30 = X
0.30 = 3/10
Therefore 3/10 of the aquarium are yellow tangs.
Answer:
15<x<27
Step-by-step explanation:
Rule for the sides of a triangle:
The sum of the two smallest sides of a triangle must be greater than the biggest side.
In this question:
Sides of 6, 21 and x. We have to find the range for x.
If 21 is the largest side:
Two smallest are 6 and x.
x + 6 > 21
x > 21 - 6
x > 15
If x is the largest side:
Two smallest and 6 and 21. So
21 + 6 > x
27 > x
x < 27
Then
x has to be greater than 15 and smaller than 27. So the answer is:
15<x<27
Hi
6% increase is multiply by 1,06
done 40 times
so : 250* 1,06^40 = 2571 ,
9514 1404 393
Answer:
a) (x³ -x² +x +2) +2/(x+1)
b) (x² +2x -5) +6/(x+3)
Step-by-step explanation:
Polynomial long division is virtually identical to numerical long division, except that the quotient term does not require any guessing. It is simply the ratio of the leading terms of the dividend and divisor. As with numerical long division, the product of the quotient term and the divisor is subtracted from the dividend to form the new dividend for the next step.
The process stops when the dividend is of lower degree than the divisor.
In part (a), you need to make sure the dividend expression has all of the powers of x present. This means terms 0x³ and 0x² must be added as placeholders in the given dividend. They will become important as the work progresses.