We have this equation.
A²+B²=C²
We have bits of information that'll help us simplify the equation so there's only one variable.
The longer leg, A, is 3 inches more than the length of the shorter leg, B, tripled.
A=3B+3
Let's plug that in.
(3b+3)²+B²=C²
The hypotenuse, C, is 3 inches less than four times the length of the shorter leg. C=4B-3
Let's plug that in.
(3b+3)²+B²=(4B-3)²
Let's solve.
9B²+18B+9+B²=16B²-24B+9
10B²+18B+9=16b²-24b+9
Let's subtract 9 from both sides.
10b²+18b=16b²-24b
Let's subtract 10b² from both sides.
18b=6b²-24b
Let's add 24b from both sides.
42b=6b²
Let's divide each side by 6.
7b=b²
With this, you can tell that b is 7 since it times 7 equal itself squared.
The shorter leg is 7 inches.
Now, let's look back at the bits of information.
The longer leg of a right triangle is 3 inches more than the length of the shorter side tripled.
3(7)+3=24
So, the longer side is 24. We can either use the other information or plug it into the equation. We can do both.
The hypotenuse is 3 less than four times the shorter leg.
4(7)-3=25
7²+24²=
49+576=625
√625=25
So, the length of the hypotenuse is 25 inches.
(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3
= a^4b^9/2a^8b^4
=b^5/2a^4
so your answer is b^5/2a^4
Answer: 3, 2, 1, 0, -1
Step-by-step explanation:
3, 2, 1, 0, -1
A) The area of the rectangle ABCD is base times height. 20in x 12in = 240 in sq.
B) The area of the triangle AED is base times height divided by two. (14in x 12in)/2 = 84in sq.
C) Figure EBCD is a right trapezoid (having one pair of parallel sides, while the others are slanted and forming a right angle).
D) Area of EBCD is the area of the rectangle minus the area of the triangle. 240in sq - 84in sq = 156in sq.
x = y^2 + 10y + 22
Divide 10 by 2 to give 5 which becaoses the second term in the parentheses
x = (y + 5)^2 - 25 + 22
x = (y + 5)^2 - 3 Answer.
