X = 3z - 22
z = 6
5x + 4y + 3z = 14
5 (3z - 22) + 4y + 3 (6) = 14
15z - 110 + 4y + 18 = 14
15 . 6 - 110 + 4y + 18 = 14
90 - 110 + 4y + 18 = 14
90 - 110 + 18 - 14 + 4y = 0
-16 + 4y = 0
4y = 16
y = 4
You take the decimal of say .48 and pull the it back so it would be 48%
Answer:
A. ![\sqrt[3]{4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%7D)
= Cube root of 4.
Step-by-step explanation:
We have been given an expression 
. We are asked to find the equivalent expression for our given expression.
Using exponent property ![\sqrt[n]{a^m} =a^{\frac{m}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D)
, we will get,
![2^{\frac{2}{3}}=\sqrt[3]{2^2}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7B2%5E2%7D)
![2^{\frac{2}{3}}=\sqrt[3]{4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7B4%7D)
Upon looking at our given choices, we can see that option A is the correct choice.
Answer:
There are (63) combinations. The notation means "six choose three". Out of six items (flavors) choose three.
(nk)=n!k!(n−k)!.
(63)=6!3!3!.
Think of it this way. There are 6 ways to choose a flavor. Once you choose, there are 5 ways to choose the next. After that, there are 4 flavors left. which is 6!/3!=6⋅5⋅4⋅3⋅2⋅13⋅2⋅1=6⋅5⋅4=120.
But, you could have chosen {chocolate,vanilla,strawberry} and you get the same combination as {vanilla, strawberry, chocolate} so we have to divide by 3!=3⋅2⋅1=6 to account for the order of choosing.
So the number of combinations of flavors is (63)=1206=20.
<h3>Mark me a brainlist</h3>
