Question

1. if df = 9x - 39, find ef.

d 47 e 3x + 10 f

Explanation:

Step1: Set up equation based on segment - addition postulate

Since $DF=DE + EF$, we have $9x - 39=47+(3x + 10)$.

Step2: Simplify the right - hand side of the equation

$9x - 39=47 + 3x+10$, so $9x - 39=3x + 57$.

Step3: Isolate the variable terms

Subtract $3x$ from both sides: $9x-3x - 39=3x-3x + 57$, which gives $6x-39 = 57$.

Step4: Isolate the variable

Add 39 to both sides: $6x-39 + 39=57 + 39$, so $6x=96$.

Step5: Solve for x

Divide both sides by 6: $\frac{6x}{6}=\frac{96}{6}$, then $x = 16$.

Step6: Find EF

Substitute $x = 16$ into the expression for $EF$ which is $EF=3x + 10$. So $EF=3\times16+10=48 + 10=58$.

Answer:

58