Answer:
x = y²
Step-by-step explanation:
Given
x =
and y = 
Note that
x =
=
×
= y²
Thus x = y²
Area of right-angled triangle is given by;
Area, A = 1/2 *b*h, Where b=base, h=height
Therefore,
A1 = 1/2bh = 16 in^2
A2 = 1/2 (2b)(2h) = 2bh
Ratio of increase = A2/A1 = {2bh}/(1/2bh} = 4 (the area is increased 4 times)
The,
A2 = 4*16 = 64 in^2
Therefore,
The area is increased by (64-16) = 48 in^2
Option C:
![$\frac{\sqrt[3]{100 x}}{5}=\sqrt[3]{\frac{4 x}{5}}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B5%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%20x%7D%7B5%7D%7D)
Solution:
Given expression is
![$\sqrt[3]{\frac{4 x}{5}}](https://tex.z-dn.net/?f=%24%5Csqrt%5B3%5D%7B%5Cfrac%7B4%20x%7D%7B5%7D%7D)
Note: ![\sqrt[3]{125}=\sqrt[3]{{5^3}} = 5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%3D%5Csqrt%5B3%5D%7B%7B5%5E3%7D%7D%20%20%3D%205)
To find the correct expression for the above simplified expression.
Option A: ![\frac{\sqrt[3]{4 x}}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B4%20x%7D%7D%7B5%7D)
5 can be written as
.
![$\frac{\sqrt[3]{4 x}}{5}=\frac{\sqrt[3]{4 x}}{\sqrt[3]{125} }](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B4%20x%7D%7D%7B5%7D%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B4%20x%7D%7D%7B%5Csqrt%5B3%5D%7B125%7D%20%7D)
![$=\sqrt[3]{\frac{4x}{125} }](https://tex.z-dn.net/?f=%24%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4x%7D%7B125%7D%20%7D)
It is not the given simplified expression.
Option B: ![\frac{\sqrt[3]{20 x}}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B20%20x%7D%7D%7B5%7D)
![$\frac{\sqrt[3]{20 x}}{5}=\frac{\sqrt[3]{20 x}}{\sqrt[3]{125} }](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B20%20x%7D%7D%7B5%7D%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B20%20x%7D%7D%7B%5Csqrt%5B3%5D%7B125%7D%20%7D)
![$=\sqrt[3]{\frac{20x}{125} }](https://tex.z-dn.net/?f=%24%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B20x%7D%7B125%7D%20%7D)
Cancel the common factor in both numerator and denominator.
![$=\sqrt[3]{\frac{4x}{25} }](https://tex.z-dn.net/?f=%24%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4x%7D%7B25%7D%20%7D)
It is not the given simplified expression.
Option C: ![\frac{\sqrt[3]{100 x}}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B5%7D)
![$\frac{\sqrt[3]{100 x}}{5}=\frac{\sqrt[3]{100 x}}{\sqrt[3]{125} }](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B5%7D%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B%5Csqrt%5B3%5D%7B125%7D%20%7D)
![$=\sqrt[3]{\frac{100x}{125} }](https://tex.z-dn.net/?f=%24%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B100x%7D%7B125%7D%20%7D)
Cancel the common factor in both numerator and denominator.
![$=\sqrt[3]{\frac{4 x}{5}}](https://tex.z-dn.net/?f=%24%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%20x%7D%7B5%7D%7D)
It is the given simplified expression.
Option D: ![\frac{\sqrt[3]{100 x}}{125}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B125%7D)
![$\frac{\sqrt[3]{100 x}}{125}=\frac{\sqrt[3]{100 x}}{5^3}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B125%7D%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B5%5E3%7D)
It is not the given simplified expression.
Hence Option C is the correct answer.
![$\frac{\sqrt[3]{100 x}}{5}=\sqrt[3]{\frac{4 x}{5}}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5Csqrt%5B3%5D%7B100%20x%7D%7D%7B5%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%20x%7D%7B5%7D%7D)
Answer:
<em>The value 60 represents the speed of the vehicle the counselor is driving.</em>
Step-by-step explanation:
<u>Linear Function</u>
The linear relationship between two variables d and t can be written in the form

Where m is the slope (or rate of change of d with respect to t) and b is the y-intercept or the point where the graph of the line crosses the y-axis
The function provided in our problem is

Where d is the distance in miles the counselor still needs to drive after t hours. Rearranging the expression:

Comparing with the general form of the line we can say m=-60, b=130. The value of -60 is the slope or the rate of change of d with respect to t. Since we are dealing with d as a function of time, that value represents the speed of the vehicle the counselor is driving. It's negative because the distance left to drive decreases as the time increases