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suter [353]
3 years ago
5

If DE = 4 x + 10, EF = 2 x -1, and DF = 9 x - 15, what is the length of DF?

Mathematics
2 answers:
maria [59]3 years ago
7 0

Answer:

DF = 57

Step-by-step explanation:

DE + EF = DF

(4x + 10) + (2x - 1) = 9x - 15

Combine like terms on the left side

6x + 9 = 9x - 15

Subtract 6x on both sides

9 = 3x - 15

Add 15 on both sides

24 = 3x

Divide by 3 on both sides

x = 8

DF = 9x - 15 = 9(8) - 15 = 72 - 15 = 57

Checking:

(4(8) + 10) + (2(8) - 1) = 9(8) - 15

(32 + 10) + (16 - 1) = 72 - 15

58 - 1 = 72 - 15

57 = 57

Maslowich3 years ago
5 0

Answer:

DF= DE+EF

9x–15 = (4x+10)+(2x-1)

9x–15 = 6x +9

9x–6x= 9+15

3x = 24

x=24/3

x=8

DF = 9x –15 = 9(8)–15= 72 –15 = 57

So; DF = 57

I hope I helped you^_^

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