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Mathematics

2 answers:

hodyreva [135]3 years ago

8 0

Answer:

Step-by-step explanation:

Veseljchak [2.6K]3 years ago

4 0

You need to substitute where the x and y are at

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3 years ago

If GF is a midsegment of CDE, find CD. A. 3.4 B. 6.8 C. 13.6 D. 14

bogdanovich [222]

Answer:

Step-by-step explanation:

Triangle CGF and triangle CED are similar. Hence, the ratio of their corresponding sides are equal. Thus we can write:

$\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}$

We can now cross multiply and solve for CD:

$\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}\\(5x+4)(3x+0.8)=(2x+3)(CD)\\15x^2+4x+12x+3.2=(2x+3)(CD)\\15x^2+16x+3.2=(2x+3)(CD)\\CD=\frac{15x^2+16x+3.2}{2x+3}$

Since GF is a midsegment of CDE, CD is double of CF. So we can write:

$CD=2CF\\\frac{15x^2+16x+3.2}{2x+3}=2(3x+0.8)\\\frac{15x^2+16x+3.2}{2x+3}=6x+1.6\\15x^2+16x+3.2=(6x+1.6)(2x+3)\\15x^2+16x+3.2=12x^2+18x+3.2x+4.8\\15x^2+16x+3.2=12x^2+21.2x+4.8\\3x^2-5.2x-1.6=0$