Answer:
yes your right
Step-by-step explanation:
not a line not a line segment and not a triangle
good job
14 tiles would be able to go on that one side of the counter
Answer:
![J = 370g](https://tex.z-dn.net/?f=J%20%3D%20370g)
Step-by-step explanation:
Given
Represent the weight of the Jar with J and the sugar with S
Initially, we have:
![J + S = 850g](https://tex.z-dn.net/?f=J%20%2B%20S%20%3D%20850g)
After 2/3 of sugar is removed we have:
![J + S -\frac{2}{3}S= 530g](https://tex.z-dn.net/?f=J%20%2B%20S%20-%5Cfrac%7B2%7D%7B3%7DS%3D%20530g)
Required
Determine the weight of the jar
--- (1)
--- (2)
Simplify (2)
![J + S -\frac{2S}{3}= 530g](https://tex.z-dn.net/?f=J%20%2B%20S%20-%5Cfrac%7B2S%7D%7B3%7D%3D%20530g)
Take LCM
![J + \frac{3S - 2S}{3}= 530g](https://tex.z-dn.net/?f=J%20%2B%20%5Cfrac%7B3S%20-%202S%7D%7B3%7D%3D%20530g)
![J + \frac{S}{3}= 530g](https://tex.z-dn.net/?f=J%20%2B%20%5Cfrac%7BS%7D%7B3%7D%3D%20530g)
Make S the subject in (1)
![J + S = 850g](https://tex.z-dn.net/?f=J%20%2B%20S%20%3D%20850g)
![S = 850g - J](https://tex.z-dn.net/?f=S%20%3D%20850g%20-%20J)
Substitute 850g - J for S in ![J + \frac{S}{3}= 530g](https://tex.z-dn.net/?f=J%20%2B%20%5Cfrac%7BS%7D%7B3%7D%3D%20530g)
![J + \frac{850g - J}{3}= 530g](https://tex.z-dn.net/?f=J%20%2B%20%5Cfrac%7B850g%20-%20J%7D%7B3%7D%3D%20530g)
Multiply through by 3
![3 * J + 3*\frac{850g - J}{3}= 530g * 3](https://tex.z-dn.net/?f=3%20%2A%20J%20%2B%203%2A%5Cfrac%7B850g%20-%20J%7D%7B3%7D%3D%20530g%20%2A%203)
![3J + 850g - J= 1590g](https://tex.z-dn.net/?f=3J%20%2B%20850g%20-%20J%3D%201590g)
Collect Like Terms
![3J - J= 1590g-850g](https://tex.z-dn.net/?f=3J%20%20-%20J%3D%201590g-850g)
![2J= 740g](https://tex.z-dn.net/?f=2J%3D%20740g)
Make J the subject
![J = \frac{1}{2} * 740g](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20740g)
![J = 370g](https://tex.z-dn.net/?f=J%20%3D%20370g)
<em>The jar weighs 370g</em>
To solve this problem you must apply the proccedure shown below:
1. You must use the formula for calculate the area of a rectangle:
![Area=(Length)(Width)](https://tex.z-dn.net/?f=Area%3D%28Length%29%28Width%29)
2. Now, you must solve for the width:
![Length=12^{\frac{1}{2} }ft} =12.5ft \\ Area=103^{\frac{1}{8} } =103.125ft^{2}](https://tex.z-dn.net/?f=Length%3D12%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7Dft%7D%20%3D12.5ft%20%5C%5C%20Area%3D103%5E%7B%5Cfrac%7B1%7D%7B8%7D%20%7D%20%3D103.125ft%5E%7B2%7D)
![Width=\frac{103.125ft^{2} }{12.5ft} =8.25ft](https://tex.z-dn.net/?f=Width%3D%5Cfrac%7B103.125ft%5E%7B2%7D%20%7D%7B12.5ft%7D%20%3D8.25ft)
Therefore, the answer is: ![8.25ft](https://tex.z-dn.net/?f=8.25ft)