Finding the area of a triangle is straightforward if you know the length of the base and the height of the triangle. But is it possible to find the area of a triangle if you know only the coordinates of its vertices? In this task, you’ll find out. Consider ΔABC, whose vertices are A(2, 1), B(3, 3), and C(1, 6); let line segment AC represent the base of the triangle. Find the area of ΔABC.

answer: 0,6

step-by-step explanation:

this would make it the y intercept.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

a

step-by-step explanation:

idk

step-by-step explanation: