How do I set up the equation and solve this problem? Thanks for any help
The seccond side of a triangle measures 5 in longer than the first side and the third side is twise as long as the seccond. The perimeter is 75 in. Find the third side.
dasgurke hat geschrieben:HI!
I hope I got your question right, because I'm not a native speaker.
Let's pretend your three sides are called (side 1) x, (side 2) z, (side 3) y.
z is given as x+5
y is given as z*2
and if I got that right, x is given as 75.
so x = 75 .
y is asked.
z = x + 5
z = 75 + 5
z = 80
y = z * 2
y = 80 * 2
y = 160
Correct me if I'm wrong
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