A and b are legs
a^2+b^2=c^2
lets say
a>b
so
a=13+3b
c=14+3b
a^2+b^2=c^2
(13+3b)^2+b^2=(14+3b)^2
9b^2+78b+169+b^2=9b^2+84b+196
10b^2+78b+169=9b^2+84b+196
minus 9b^2 both sides
b^2+78b+169=84b+196
minus 84b both sides
b^2-6b+169=196
minus 196 both sides
b^2-6b-27=0
factor
(b+3)(b-9)=0
set to zero
b+3=0
b=-3, false, dimentions cannot be negative
b-9=0
b=9
shorter leg is 9
a=13+3b
a=13+3(9)
a=13+27
a=40
c=14+3b
c=14+27
c=41
side legnths are
9in, 40in, 41in
Answer:
P’(-1, 1), I’ (1, 2), G’(1,0)
Step-by-step explanation:
Plug in -2 for a, 3 for b, and -5 for c. Then it'd be | (-3)^2 - 2(-2)(-5) +5 (3) |. Then, multiply everything to get | 9 -20 +15 |. Then, add/subtract to get | 4 |. Finally, take the absolute value and get 4.
Answer:
X=10, x=4
Step-by-step explanation:
first, divide 14 by 2, and square it; giving you
x^2-14x+(7)^2=-40
Because 7^2 is 49, you have to add 49 to the other side; giving you
x^2-14x+(7)^2=9
Then factor the left side; giving you (x-7)^2=9
Take the square root of both sides, giving you
(x-7)^2= plus or minus 3
Then solve to get two solutions, 10 and 4
