Answer:
The lengths 7, 40 and 41 cannot be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.
Step-by-step explanation:
You can test this by using the Pythagorean theorem, which states a^2+b^2=c^2
with 7,40 and 41
7^2+40^2 does not equal to 41^2
but with 12,16, and 20
12^2+16^2=20^2
therefore, the last statement is correct.
Answer:
Step-by-step explanation:
find the ration for the valid number
(13x+14)(6x-5)=0
78x²-65x+84x-70=0
78x²+19x-70=0
It was equation.
______________
Factoring⇒ a×c=78×(-70)=-5460
84×(-65)=-5460
84+(-65)=19 which is b
__________________
78x²+84x-65x-70=0
6x(13x+14)-5(13x+14)=0
(13x+14)(6x-5)=0
13x+14=0 6x-5=0
13x=-14 6x=5
x=-14/13 x=5/6
Two solution X=-14/13;5/6
Answer:
7
−
2
(
−
1
0
)
=
4
0
7x{\color{#c92786}{-2(x-10)}}=40
7x−2(x−10)=40
7
−
2
+
2
0
=
4
0
Step-by-step explanation: therefore your answer would be x=4
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