I think it would be A. 9.6. I'm completely sure on that one.
11. You’ve done it correctly
12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2
13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution
14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath
15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4
Hope this helped :)
Answer:
(x, y) = (2, 9)
Step-by-step explanation:
For the triangles to be congruent, the hypotenuses must be the same length:
y = x + 7
and the marked leg must be the same length in each triangle:
y -3 = 4x -2
These are two equations in two unknowns (a "system" of equations) that can be solved in any of the usual ways. Since the first equation gives an expression for y, it is convenient to substitute that into the second equation:
(x +7) -3 = 4x -2
x +4 = 4x -2 . . . . . . collect terms
x +6 = 4x . . . . . . . . .add 2
6 = 3x . . . . . . . . . . . subtract x
2 = x . . . . . . . . . . . . divide by 3
y = 2 + 7 = 9 . . . . . .substitute for x in the first equation
The values you're looking for are x = 2, y = 9.
