Triangle DEF has vertices located at D (2, 1), E (3, 5), and F (6, 2). Part A: Find the length of each side of the triangle. Show your work. (4 points)

Part B: Find the slope of each side of the triangle. Show your work. (3 points)

Part C: Classify the triangle. Explain your reasoning. (3 points)

Step-by-step explanation:

D(2, 1) E(3, 5) F(6, 2)

Part A

use the distance formula d= √(x2-x1)² +(y2-y1)²

DE = √(2-3)²+(1-5)² = √1 +16 = √17

EF= √(3-6)² +(5-2)² = √9 +9 = √18= √2*3*3= 3√2

DF = √(2-6)² +(1-2)² = √16 + 1= √17

Part B

slope is m= (y1-y1) / (x2-x1)

m of DE = (1-5) / (2-3) = -4/-1 = 4

m of EF =  (5-2) / (3-6)= 3/-3 = -1

m of DF =  (1-2) / (2-6) = -1/-4 = 1/4

Part C

the triangle has sides DE = DF = √17 and DE⊥DF because their slopes are negative reciprocals ( 4 and -1/4) so the triangle is an isosceles right triangle

i'm not entirely sure

step-by-step explanation:

if you ask home depot or lowe's they might be able to you out

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2