The sales of digital cameras (in millions of units) in year t is given by the function
f(t) = 3.06t + 6.84 (0 ≤ t ≤ 3)
where t = 0 corresponds to the end of 2001. over that same period, the sales of film cameras (in millions of units) is given by
g(t) = −1.85t + 16.48 (0 ≤ t ≤ 3).
(a) show that more film cameras than digital cameras were sold in 2001.
digital cameras
million
film cameras
million

(b) when did the sales of digital cameras first exceed those of film cameras?

Step-by-step explanation:

The function representing the sales of digital cameras (in millions of units) in year t is

f(t) = 3.06t + 6.84 (0 ≤ t ≤ 3)

Also,

The function representing the sales of film cameras (in millions of units) is given by

g(t) = −1.85t + 16.48 (0 ≤ t ≤ 3)

a) where t = 0 corresponds to the end of 2001,

The number of digital cameras sold in 2001 would be

3.06 × 0 + 6.84 = 6.84 digital cameras.

The number of film cameras sold in 2001 would be

−1.85 × 0 + 16.48 = 16.48 film cameras.

Therefore, more film cameras than digital cameras were sold in 2001.

b) 3.06t + 6.84 = −1.85t + 16.48

3.06t + 1.85t = 16.48 - 6.84

4.91t = 9.64

t = 9.64/4.91 = 1.96

Therefore, the sales of digital cameras first exceed those of film cameras at t = 2 and that is in 2003.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy