yes, they are all proportional
We know that
the equation of the parabola is of the form
y=ax²+bx+c
in this problem
y=1/4x²−x+3
where
a=1/4
b=-1
c=3
the coordinates of the focus are
(-b/2a,(1-D)/4a)
where D is the discriminant b²-4ac
D=(-1)²-4*(1/4)*3-----> D=1-3---> D=-2
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> 2
y coordinate of the focus
(1-D)/4a------> (1+2)/(4/4)---> 3
the coordinates of the focus are (2,3)
5+3+-9+3=2 and then the varibles so the answer is 2 cubed
Answer:
9.
h(x)=-x-1
let y =-x-1
interchanging role of x & y
x=-y-1
y=-x-1
h-¹(x)=-x-1
again
f(x)=(2-3x)/2
let
y=(2-3x)/2
interchanging role of x & y
x=(2-3y)/2
2x-2=-3y
y=(2-2x)/3
f-¹(x)=(2-2x)/3
<u>G</u><u>i</u><u>v</u><u>e</u><u>n</u><u> </u><u>f</u><u>u</u><u>n</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>n</u><u>o</u><u>t</u><u> </u><u>i</u><u>n</u><u>v</u><u>e</u><u>r</u><u>s</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u>e</u><u>a</u><u>c</u><u>h</u><u> </u><u>o</u><u>t</u><u>h</u><u>e</u><u>r</u><u>.</u>
10.
f(x)=-x+3
let y=-x+3
interchanging role of x & y
x=-y+3
y=3-x
f-¹(x)=-x+3
equal to g(x)=-x+3
<u>Given function are</u><u> inverse of each other.</u>
I think it’s 18/5 maybe idk try it!