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Function A

f(x)  =  1 / ( x - 3)

The  vertical asymptote is  the value of x that makes x - 3 = 0 ⇒   x = 3

The vertical asymptote of B   is  x = 4

So....

Function A has a vertical asymptote at x = 3.
Function B has a vertical asymptote at x = 4.

  W*F 

                                                                                                        -_-

Nice, EP  !!!

Diameter = 8.5 cm

So

Radius = 4.25

V_cone =   3.14 * (radius)^2 * height /3  =   3.14 * (4.25)^2 * 4.5  / 3   ≈  85.07 cm^3 

f(x)  =  sqrt (-x)

Because the quantity under the root must be greater or equal to 0....then...

The domain of A is (-inf, 0 ]

The domain of B =  g(x)  is exactly the same

So

C.) The domain of both functions is the set of real numbers less than or equal to 0.

Draw this on a graph yourself.

draw a line through AA'

draw a line through BB'

draw a line through CC'

where they all cross is the centre of the dilation. Let the centre be X

For the scale, take the distance of the new point (any new point) from the centre and divide it by the distance of the equavalent old point from the centre.

eg       XA' divided by XA     is the scale factor.

Function A has two x intercepts  since the points (3, 0) and (1,0)  are on the graph

To find out the number  of x intercepts that B has

l x l - 2 = 0

l x l = 2

Two values of x satify this   x = 2  and x = 2

So....B has two x intercepts

So

Both functions have the same number of x-intercepts.

[ But not the SAME x intercepts ]  !!!!

This is just a substitution question.

$x_{n+1}=(x_n)^2-x_n\\ so\\ x_{2}=(x_1)^2-x_1\\ If \;\; x_1=1\\ then\; x_{2}=(1)^2-1=0\\ $

You can do the others yourself!

Do your own homework!

Ask one question - after you have made a serious attempt on your own!

5kx + 6 = 7kx 

Subtract 6 on both sides 

5kx + 6 - 6 = 7kx - 6

5kx = 7kx - 6

Then subtract 7kx on both sides

5kx - 7kx = 7kx - 6 - 7kx 

-2kx = -6 

Now divide by -2k on both sides

-2kx/-2k = -6/-2k

x = -6/-2k

Simplify

x = 3/k

The answer is D 

These are algebriac equations that you keep doing. Maybe try practicing the simpiler algebriac equations then slowely do more complicated ones. Just letting you know. 

Hope this helps ;P