rationalizing the numerator, or namely, "getting rid of that pesky radical at the top".
we simply multiply top and bottom by a value that will take out the radicand in the numerator.
![\bf \cfrac{\sqrt[3]{144x}}{\sqrt[3]{y}}~~ \begin{cases} 144=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\\ \qquad 2^3\cdot 18 \end{cases}\implies \cfrac{\sqrt[3]{2^3\cdot 18x}}{\sqrt[3]{y}}\implies \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}} \\\\\\ \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}\cdot \cfrac{\sqrt[3]{(18x)^2}}{\sqrt[3]{(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)(18x)^2}}{\sqrt[3]{(y)(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)^3}}{\sqrt[3]{18^2x^2y}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B144x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D~~%0A%5Cbegin%7Bcases%7D%0A144%3D2%5Ccdot%202%5Ccdot%202%5Ccdot%202%5Ccdot%203%5Ccdot%203%5C%5C%0A%5Cqquad%202%5E3%5Ccdot%2018%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B%2818x%29%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%2818x%29%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%2818x%29%2818x%29%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%28y%29%2818x%29%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%2818x%29%5E3%7D%7D%7B%5Csqrt%5B3%5D%7B18%5E2x%5E2y%7D%7D)
![\bf \cfrac{2(18x)}{\sqrt[3]{324x^2y}}~~ \begin{cases} 324=2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3\\ \qquad 12\cdot 3^3 \end{cases}\implies \cfrac{36x}{\sqrt[3]{12\cdot 3^3x^2y}} \\\\\\ \cfrac{36x}{3\sqrt[3]{12x^2y}}\implies \cfrac{12x}{\sqrt[3]{12x^2y}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%2818x%29%7D%7B%5Csqrt%5B3%5D%7B324x%5E2y%7D%7D~~%0A%5Cbegin%7Bcases%7D%0A324%3D2%5Ccdot%202%5Ccdot%203%5Ccdot%203%5Ccdot%203%5Ccdot%203%5C%5C%0A%5Cqquad%2012%5Ccdot%203%5E3%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B36x%7D%7B%5Csqrt%5B3%5D%7B12%5Ccdot%203%5E3x%5E2y%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B36x%7D%7B3%5Csqrt%5B3%5D%7B12x%5E2y%7D%7D%5Cimplies%20%5Ccfrac%7B12x%7D%7B%5Csqrt%5B3%5D%7B12x%5E2y%7D%7D)
A polynomial function of least degree with integral coefficients that has the
given zeros 
Given
Given zeros are 3i, -1 and 0
complex zeros occurs in pairs. 3i is one of the zero
-3i is the other zero
So zeros are 3i, -3i, 0 and -1
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
the factor form of given zeros

Now we multiply it to get the polynomial

polynomial function of least degree with integral coefficients that has the
given zeros 
Learn more : brainly.com/question/7619478
Answer:
There is an 84.97% probability that at least six wear glasses.
Step-by-step explanation:
For each adult over 50, there are only two possible outcomes. Either they wear glasses, or they do not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
In this problem we have that:

What is the probability that at least six wear glasses?

There is an 84.97% probability that at least six wear glasses.
Answer:
2x+4
Step-by-step explanation:
Distribute:
=(2)(x)+(2)(−6)+(2)(8)
=2x+−12+16
Combine Like Terms:
=2x+−12+16
=(2x)+(−12+16)
=2x+4