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.

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15