In ΔHIJ, the measure of ∠J=90°, the measure of ∠H=71°, and JH = 2 feet. Find the length of IJ to the nearest tenth of a foot.
1 answer:
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Easy, showing work,
using the method FOIL (First,Outer,Inner, Last)
Take, (x+3)(x+5) and multiply,
First,
x*x=
Outer,
x*5=5x
Inner,
3*x=3x
Last,
3*5=15
Add the terms together,
+8x+15 (+5x+3x+15)
Thus, your answer.
using the method FOIL (First,Outer,Inner, Last)
Take, (x+3)(x+5) and multiply,
First,
x*x=
Outer,
x*5=5x
Inner,
3*x=3x
Last,
3*5=15
Add the terms together,
Thus, your answer.