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20. AB = 42

21. BC = 28

22. AC = 70

23. BC = 20.4

24. FH = 48

25. DE = 10, EF = 10, DF = 20

Step-by-step explanation:

✍️Given:

AB = 2x + 7

BC = 28

AC = 4x,

20. Assuming B is between A and C, thus:

AB + BC = AC (Segment Addition Postulate)

2x + 7 + 28 = 4x (substitution)

Collect like terms

2x + 35 = 4x

35 = 4x - 2x

35 = 2x

Divide both side by 2

17.5 = x

AB = 2x + 7

Plug in the value of x

AB = 2(17.5) + 7 = 42

21. BC = 28 (given)

22. AC = 4x

Plug in the value of x

AC = 4(17.5) = 70

✍️Given:

AC = 35 and AB = 14.6.

Assuming B is between A and C, thus:

23. AB + BC = AC (Segment Addition Postulate)

14.6 + BC = 35 (Substitution)

Subtract 14.6 from each side

BC = 35 - 14.6

BC = 20.4

24. FH = 7x + 6

FG = 4x

GH = 24

FG + GH = FH (Segment Addition Postulate)

(substitution)

Collect like terms

Divide both sides by -3

FH = 7x + 6

Plug in the value of x

FH = 7(6) + 6 = 48

25. DE = 5x, EF = 3x + 4

Given that E bisects DF, therefore,

DE = EF

5x = 3x + 4 (substitution)

Subtract 3x from each side

5x - 3x = 4

2x = 4

Divide both sides by 2

x = 2

DE = 5x

Plug in the value of x

DE = 5(2) = 10

EF = 3x + 4

Plug in the value of x

EF = 3(2) + 4 = 10

DF = DE + EF

DE = 10 + 10 (substitution)

DE = 20

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

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step-by-step explanation: