First, draw a diagram to visualize the situation:
As we can see, the difference between the x-coordinates of the points is the length of one leg of a right triangle, and the difference between the y-coordinates is the length of the other leg.
Then, for the given points (-5,4) and (8,-3), the distance given by the Pythagorean Theorem is:

Therefore, the distance between the points (-5,4) and (8,-3) is: √218.
10 quarts = 2.5 gallons.
There are 4 quarts for every 1 gallon, but because you're finding gallons and not quarts, you do 10 ÷ 4 = 2.5.
Answer:
The simplified form of the expression is ![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step-by-step explanation:
Given : Expression ![7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Csqrt%5B3%5D%7B16x%7D-3%5Csqrt%5B3%5D%7B8x%7D)
To Simplified : The expression
Solution :
Step 1 - Write the expression
![7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Csqrt%5B3%5D%7B16x%7D-3%5Csqrt%5B3%5D%7B8x%7D)
Step 2- Simplify the roots and re-write as
and 
![7\sqrt[3]{2x}-3\times2\sqrt[3]{2x}-3\times2\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7Bx%7D)
Step 3- Solve the multiplication
![7\sqrt[3]{2x}-6\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step 4- Taking
common from first two terms
![\sqrt[3]{2x}(7-6)-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%287-6%29-6%5Csqrt%5B3%5D%7Bx%7D)
![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Therefore, The simplified form of the expression is ![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
2 over 3 is 14 over 21 simplified, dividing both numbers by 7.
The average rate of change over a given interval is the slope of the secant line through the endpoints of the interval.
$203/unit.The instantaneous rate of change at a certain point is the slope of the tangent line.
We differentiate C to get 2x. We substitute 100 for x to get that the instantaneous rate of change at 100 is
$200/unit.