Answer:
Third choice from the top is the one you want
Step-by-step explanation:
This whole concept relies on the fact that if the index of a radical exactly matches the power under the radical, both the radical and the power cancel each other out. For example:
and another example:
![\sqrt[12]{2^{12}}=2](https://tex.z-dn.net/?f=%5Csqrt%5B12%5D%7B2%5E%7B12%7D%7D%3D2)
Let's take this step by step. First we will rewrite both the numerator and the denominator in rational exponential equivalencies:
![\frac{\sqrt[4]{6} }{\sqrt[3]{2} }=\frac{6^{\frac{1}{4} }}{2^{\frac{1}{3} }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B6%7D%20%7D%7B%5Csqrt%5B3%5D%7B2%7D%20%7D%3D%5Cfrac%7B6%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%7D%7B2%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%7D)
In order to do anything with this, we need to make the index (ie. the denominators of each of those rational exponents) the same number. The LCM of 3 and 4 is 12. So we rewrite as

Now we will put it back into radical form so we can rationalize the denominator:
![\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B12%5D%7B6%5E3%7D%20%7D%7B%5Csqrt%5B12%5D%7B2%5E4%7D%20%7D)
In order to rationalize the denominator, we need the power on the 2 to be a 12. Right now it's a 4, so we are "missing" 8. The rule for multiplying like bases is that you add the exponents. Therefore,

We will rationalize by multiplying in a unit multiplier equal to 1 in the form of
![\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B12%5D%7B2%5E8%7D%20%7D%7B%5Csqrt%5B12%5D%7B2%5E8%7D%20%7D)
That looks like this:
![\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }*\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B12%5D%7B6%5E3%7D%20%7D%7B%5Csqrt%5B12%5D%7B2%5E4%7D%20%7D%2A%5Cfrac%7B%5Csqrt%5B12%5D%7B2%5E8%7D%20%7D%7B%5Csqrt%5B12%5D%7B2%5E8%7D%20%7D)
This simplifies down to
![\frac{\sqrt[12]{216*256} }{\sqrt[12]{2^{12}} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B12%5D%7B216%2A256%7D%20%7D%7B%5Csqrt%5B12%5D%7B2%5E%7B12%7D%7D%20%7D)
Since the index and the power on the 2 are both 12, they cancel each other out leaving us with just a 2! Doing the multiplication of those 2 numbers in the numerator gives us, as a final answer:
![\frac{\sqrt[12]{55296} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B12%5D%7B55296%7D%20%7D%7B2%7D)
Phew!!!
Answer:
the ladder reaches 8 feet up
Step-by-step explanation:
:D
Y= -5x -4 because you would do slope intercept form and sub the y and x in for -4 and 0
Answer:
x = 2, y = 4
Step-by-step explanation:
x +
y = 5 ( multiply through by 4 to clear the fraction )
4x + 3y = 20 → (1)
x -
y = 0 ( multiply through by 2 to clear the fraction )
2x - y = 0 → (2)
Multiplying (2) by 3 and adding to (1) will eliminate y
6x - 3y = 0 → (3)
Add (1) and (3) term by term to eliminate y
10x + 0 = 20
10x = 20 ( divide both sides by 10 )
x = 2
Substitute x = 2 into either of the 2 equations and solve for y
Substituting into (1)
4(2) + 3y = 20
8 + 3y = 20 ( subtract 8 from both sides )
3y = 12 ( divide both sides by 3 )
y = 4
Solution is x = 2, y = 4