ALGEBRA II // MODULE 1 FUNCTIONS AND INVERSES - 1.2
1.2
READY, SET, GO!
Name
Period
Date
READY
Topic: Solving for a variable
Solve for x.
1.
17 = 5x + 2
2. 2x² – 5 = 3x2 – 12x +
31
3. 11 = 2x + 1

1. x = 3

2.x = 6

3. x = 5

I hope this helps you :D

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²