THESE ARE PROBABLY WROMG but help me please i am gonna sob

Problem 5

Point B is the midpoint of AC. This means B cuts AC in half to get the smaller pieces AB and BC which are equal to one another.

AB = BC

AC = AB+BC

AC = AB+AB

AC = 2x+2x

AC = 4x

Because AC = 20 as well, this means,

4x = 20

x = 20/4

x = 5

So, AB = 2x = 2*5 = 10 and BC is this length as well.

Problem 6

Use the segment addition postulate.

RS + ST = RT

x + 3x = 24

4x = 24

x = 24/4

x = 6

RS = x = 6

ST = 3x = 3*6 = 18

Problem 7

We'll use the same idea as problem 5.

DE = EF

3x+10 = 8x-5

10+5 = 8x-3x

15 = 5x

5x = 15

x = 15/5

x = 3

So

DE = 3x+10 = 3*3+10 = 9+10 = 19

EF = 8x-5 = 8*3-5 = 24-5 = 19

DF = DE+EF = 19+19 = 38

Problem 8

This problem is similar to problem 6. We use the segment addition postulate.

GH + HI = GI

3x + 10 = 40

3x = 40-10

3x = 30

x = 30/3

x = 10

GH = 3x = 3*10 = 30

Problem 9

The double tickmarks indicate the segments KL and LM are equal in length. That means point L bisects KM.

KL = LM

KL + LM = KM

x+x = 25

2x = 25

x = 25/2

x = 12.5

a function could be identified in 3 ways either on a chart,graph, or ordered pairs. first a function is that each point has its own. the rule is that there can be many x's to one y but there can not be 2 x's at different y's. so on a graph when you put a straight line through the y's there should be one dot only to be a function.

step-by-step explanation:

7.3

step-by-step explanation: