Find the perimeter of triangle ABC with vertices A(6, 3), B(6, - 2) , and C(- 4, 3) .

26.2 units

Step-by-step explanation:

We are given the points/vertices

A(6, 3),

B(6, - 2) , and

C(- 4, 3)

Step two

Let us find the distances between the given points/vertices

A-B =A(6, 3) to B(6,-2)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6-6)²+(-2-3)²)

d=√(-2-3)²)

d=√(-5)²)

d=5 units

B-C=B(6, - 2) to C(-4, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((-4-6)²+(3+2)²)

d=√(-10)²+(5)²)

d=√100+25

d=√125

d=11.2 units

C-A=C(-4, 3) to A(6, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6+4)²+(3-3)²)

d=√(10)²

d=√100

d=10 units

Hence the perimeter is 5+11.2+10

P=26.2 units

answer is  

c.

36

step-by-step explanation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: