34. Suppose that g is a function from A to B and f is a function from B to C. Prove each of these statements. a) If f ◦g is onto, then f must also be onto. b) If f ◦g is one-to-one, then g must also be one-to-one. c) If f ◦g is a bijection,

Step-by-step explanation:

Given that g is a function from A to B and f is a function from B to C.

g: A -->B  and f: B-->C

a) fog= f{g[x)} is a function from A to C

Let fog be onto.  Then we get for any element C in f we got an image in A.

This is possible only if every element of C has an image in B because if not then f cannot be applied to g(x).

Hence proved

b) Let fog be one to one.  

i.e. fog(a) = fog(b) implies a =b

This suggests that f{g(a) }=f{g(b)}

Since f is a function for g(a) and g(b) three exists unique images.

So g(a) = g(b) implies a = b.

g is one to one

c) If fog is one to one and on to then we have f is onto and g is one to one.

Since f is onto every element in g has a preimage so it follows that both f and g are bijective.

a. $250

b. The total cost for x month is $200 + $50 × x

c. Please see the attached graph

d. Yes, there is a proportional relationship between time and the cost of the cell phone plan

e. $350

Step-by-step explanation:

The given parameters are;

The costs to start on the cell phone cell = $200

The charge each month = $50

a. The total cost to use the cell phone plan for 1 month = Start up fee + Monthly charge = $200 + $50 = $250

b. The total cost for x month, c = $200 + $50 × x

c. Please see the attached graph

d. Yes, there is a proportional relationship between time and the cost of the cell phone plan

e. The line parallel to the line graphed that goes to the point (0, 350), is given as follows;

y = m·x + c

Where;

m = 50

y - 350 = 50 × (x - 0)

y = 50·x + 350

The startup fee is given by the total cost at x = 0, when the monthly fee is excluded, which gives;

c = 50 × 0 + 350 = 350

The startup fee for the pricing plan = $350.