A = s^2
A = 40
40 = s^2...by taking the square root of both sides, u eliminate the ^2
sqrt 40 = s
6.32 = s <===
Answer:
x=5
Step-by-step explanation:
Since this is a right triangle, we can use the pythagorean theorem
a^2 + b^2 = c^2
x^2+ ( 2x+2)^2 = (2x+3)^2
FOIL the squared terms
x^2 +(4x^2 + 4x+4x+4) = 4x^2 + 6x+6x +9
Combine like terms
5x^2+8x +4 = 4x^2 +12x +9
Subtract 4x^2 +12x +9 from each side
5x^2+8x +4 -4x^2 -12x -9= 4x^2 +12x +9-4x^2 -12x -9
x^2 -4x -5 =0
Factor
(x-5) (x+1) = 0
Using the zero product property
x=5 x=-1
But x cannot be negative or we would have negative length
x=5
Step-by-step explanation:
Change it into this
(2x-3)(2x-3)
Then multiply the insides and outsides and add them
(2x*2x) + (2x* -3) + (-3*2x) + (-3*-3)
Simplify
4x^2 - 6x -6x + 9
Simplify again
4x^2 - 12x + 9
