Answer:
37.5 square units
Step-by-step explanation:
Given:
A(-6,-4), B(6,5), C(-1,6), and D(2, 2), where CD is the altitude if ∆ABC
Required:
Area of ∆ABC
SOLUTION:
Area of a ∆ABC =½*AB*CD
✍️
✍️
✍️Area of a ∆ABC =½*AB*CD
= ½*15*5
✅ Area = 37.5 square units
Answer:
.
Step-by-step explanation:
Let the 
-coordinate of 
be 
. For 
to be on the graph of the function 
, the 
-coordinate of 
would need to be 
. Therefore, the coordinate of 
would be 
.
The Euclidean Distance between and 
is:
.
The goal is to find the a that minimizes this distance. However, 
is non-negative for all real 
. Hence, the 
that minimizes the square of this expression, 
, would also minimize 
.
Differentiate with respect to :
![\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5Bt%5E2%20-%2017%20%5C%2C%20t%20%2B%2081%5Cright%5D%20%3D%202%5C%2C%20t%20-%2017)
.
![\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E%7B2%7D%7D%7Bdt%5E%7B2%7D%7D%5Cleft%5Bt%5E2%20-%2017%20%5C%2C%20t%20%2B%2081%5Cright%5D%20%3D%202)
.
Set the first derivative, 
, to 
and solve for :
.
.
Notice that the second derivative is greater than for this . Hence, would indeed minimize . This value would also minimize , the distance between and .
Therefore, the point would be closest to when the -coordinate of is .
We apply the following property:
(x^n)^m = x^(n • m)
Let m = 2
(P^1/2)^2 = P^(1/2 • 2) = P
Answer:
1st 2nd or both's answer you want
Answer:
56
Step-by-step explanation:
