Answer:
Distance of JK = 15 unit
Step-by-step explanation:
Given:
J(4,8)
K(-1,-2)
Find:
Distance of JK
Computation:
Distance = √(x₂-x₁)² + (y₂-y₁)²
Distance of JK = √(-1-4)² + (-2-8)²
Distance of JK = √25 + 100
Distance of JK = √125
Distance of JK = 15 unit
Answer:
not sure
Step-by-step explanation: not good at this stuf
Step-by-step explanation:
![\frac{y_2-y_1}{x_2-x_1}=\frac{15-(-13)}{28-(-28)}\\=\frac{28}{2(28)}\\\therefore\ m=\frac{1}{2}\\\frac{y-y_1}{xl-x_1}=m]\\\frac{y+13}{x+28}=\frac{1}{2}\\2y+26=x+28\\2y=x+2\\ y=\frac{1}{2}x+1](https://tex.z-dn.net/?f=%20%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%3D%5Cfrac%7B15-%28-13%29%7D%7B28-%28-28%29%7D%5C%5C%3D%5Cfrac%7B28%7D%7B2%2828%29%7D%5C%5C%5Ctherefore%5C%20m%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C%5Cfrac%7By-y_1%7D%7Bxl-x_1%7D%3Dm%5D%5C%5C%5Cfrac%7By%2B13%7D%7Bx%2B28%7D%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C2y%2B26%3Dx%2B28%5C%5C2y%3Dx%2B2%5C%5C%20y%3D%5Cfrac%7B1%7D%7B2%7Dx%2B1)
In order to find y for point C on AB, substitute point C in line equation if AB.
Since 1/4 is equal to 0.25 than you would do 5,400 times 0.25 and get the answer of 13,500.0. So the race car weighs 13,500 pounds.
