<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)
The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem
⇔
units
Therefore the perpendicular distance from B to AC is =
This is what I got hope it helps!
Answer:
Probability (bid accepted) = 0.48
Step-by-step explanation:
Probability density is given byF(y)= 1/(b-a)
a=9500
b= 14700
F(y)= 1/(14700-9500) =1/5200=0.00019
Probability (bid accepted)= (12000-9500)÷1/5200
P( bid accepted) = 2500×0.00019=0.475 approximately 0.48
Because Philip is conducting a survey that pertains to the weight of the cows, he needs statistics that directly relate to weight.
The cows being overweight doesn't give Philip concrete information on the weight of the cows, as it doesn't give the amount of pounds a cow weighs. Average age has nothing to do with weight, and normal body weight doesn't give concrete information on weight.
The answer is "<span>What is the weight of each cow on the farm?".</span>
