Let;
A(-8,6) B(6,6) C(6, -4) D(-8, -4)
Let's find the length AB
x₁= -8 y₁=6 x₂=6 y₂=6
We will use the distance formula;
![d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![=\sqrt[]{(6+8)^2+(6-6)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%286%2B8%29%5E2%2B%286-6%29%5E2%7D)
![=\sqrt[]{14^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%2B0%7D)
Next, we will find the width BC
B(6,6) C(6, -4)
x₁= 6 y₁=6 x₂=6 y₂=-4
substitute into the distance formula;
![d=\sqrt[]{(6-6)^2+(-4-6)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%286-6%29%5E2%2B%28-4-6%29%5E2%7D)
![=\sqrt[]{(-10)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-10%29%5E2%7D)
![=\sqrt[]{100}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B100%7D)
Area = l x w
= 14 x 10
= 140 square units
15 is the answer because 8-5 is 3 and 3 and 4 is 12 so 12 plus 3 is 15
The other factor is (y - 3). If you multiply the two terms together, you will end up with
:
Answer:
c) none of the above
Step-by-step explanation:
this is because we should put the equation like:
1/3 - (-4/3) because the distance is 5 units and
1/3 + 4/3 = 5/3
this is because (-) × (-) = (+)
This is not a function. There are multiple outputs for the same input at 
. A function only has 1 output for every input.
