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Can someone help me with this one? It’s very difficult too me

Can someone help me with this one? It’s very difficult too me-example-1
User Kemzie
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7.8k points

2 Answers

6 votes

Answer:

The year 2019.

Explanation:

Plug 15,000 into the variable C:

15,000 = 20t^2 + 135t + 3050

20t^2 + 135t - 11,950 = 0. Divide through by 4:

4t^2 + 27t - 2390 = 0.

t = [ (-27 +/- sqrt (27^2 - 4 * 4 * -2390)] / (2*4)

= 21.3, -28.05 ( we ignore the negative value).

So the number of cars will reach 15,000 in 1998 + 21 = 2019.

User Kishor Velayutham
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7.7k points
4 votes

The equation is C = 20t^2 + 135t + 3050

You are told the total number of cars sold is 15000.

Replace c with 15,000 and solve for t:

15000 = 20t^2 + 135t + 3050

Subtract 15000 from both sides:

0 = 20t^2 + 135t - 11950

Use the quadratic formula to solve for t.

In the quadratic formula -b +/-√(b^2-4(ac) / 2a

using the equation, a = 20, b = 135 and c = -11950

The formula becomes -135 +/- √(135^2 - 4(20*-11950) / (2*20)

t = 21.3 and -28.1

T has to be a positive number, so t = 21.3,

Now you are told t = 0 is 1998,

so now add 21.3 years to 1998

1998 + 21.3 = 2019.3

So in the year 2019 the number of cars will be 15000

User Jes
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7.3k points