Answer:
By hypotenuse - side test (HL) the two triangles are congruent.
Step-by-step explanation:
In ∆ABC and ∆DCB
i) angleABC = angleDCB.....(each 90°)
ii) BC = BC .....( common side)
iii) AC = DB.....(given)
therefore by hypo-side test ∆ABC congruent ∆DCB
The awenser of the problem is A
a(x+1)(x-1)+b(x-2)(x+1)+(x+1)^2=9x^2-x-10
(x+1)(a(x-1)+b(x-2)+(x+1))=(x+1)(9x-10)
a(x-1)+b(x-2)+x+1=9x-10
Now this equation is much simpler!
(a+b)x-a-2b+x+1=9x-10
(a+b)x-a-2b=8x-11
(a+b-8)x-a-2b-11=0
a+b-8=(a+2b-11)/x
I can't solve it 3 variables and 1 equations means infinite answers so yea.
Answer:
b
Step-by-step explanation:
add 5+3 then add 1/2 plus 3/4
