The way to distribute the eggs at the lowest cost would be using 5 cartons of 12 eggs and 1 carton of 6 eggs.
Since Timmy is helping his grandpa on his chicken farm, and he has to pack 66 eggs into cartons, minimizing the packing cost, and he can use 12-egg cartons and 6-egg cartons, and two 12-egg cartons cost as much as three 6-egg cartons, to determine how many cartons of each should be used, the following calculation must be performed:
- 2 x 12 = 3 x 6
- 24 = 18
- Thus, assuming as an example that a total of 2 cartons of 12 eggs and 3 cartons of 6 eggs cost $ 3, the value of each carton would be the following:
- 3/3 = 1
- 3/2 = 1.5
- 66/12 = 5.5
- 5 x 1.5 + 1 x 1 = X
- 7.5 + 1 = X
- 8.5 = X
In this way, the way to distribute the eggs at the lowest cost would be using 5 cartons of 12 eggs and 1 carton of 6 eggs.
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Divide 2/3 by 3/64 and you will get the number of cups of punch. It doesn't come out evenly, but that's the way to do this.there is a total of 2/3 gallon of punch. Each guests has a cup of 3/64 gallon of punch. Now many 3/64 gallon of punch can you fit in 2/3 gallon of punch, that will tell you how many guests there are. so naturally we want to divide<span> ( 2/3 ) / ( 3/64).</span>
Answer:
x = ± 2
Step-by-step explanation:
Given
x² + 5 = 3x² - 3 ( subtract x² + 5 from both sides )
0 = 2x² - 8 ( add 8 to both sides )
8 = 2x² ( divide both sides by 2 )
4 = x² ( take the square root of both sides )
x = ±
= ± 2
Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;


Divide by m to isolate
.


To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)