Question

Help!!!

Answer 1

Answer:

2019

Step-by-step explanation:

first you should change the statement into quadratic equation and replace c with 15000.

c=20t^2+135t+3050.

15000=20t^2+135t+3050

15000-3050=20t^2+135t

11950=20t^2+135t. then write the equation in standard quadratic form.

20t^2+135t-11950=0. after this you got 4 ways of solving the quadratic equation but I am just gone using quadratic formula:

-b+ - √b^2-4ac. a,b and C stand for the

2a. the coefficients

-135+ - √((135)^2-4(20)(-11950))

2(20)

-135+ - √(974225))

)) 40

-135 - 987 or -135+ 987

40. 40

-1122/40. or 852/40

- 28.05 or. 21.3

in this case we have to answer but time cannot be negative we take value 21.3(it is the approximate value)

so we add 21.3 to 1998 to find the year

>>21.3 + 1998 = 2019.3 but we write it as 2019 instead of 2019.3

Answer 2

Answer:

In the year 2019 the number of new cars purchased will reach 15,000.

Step-by-step explanation:

t = 0 corresponds to the number of new cars purchased in 1998. If that is so, we can determine t ( time ) by making our quadratic equation here equal to 15,000 - considering that we want the year the number of cars reaches this value. t here is only the number of years to reach 15,000 cars, so we would have to add that value to 1998, to see the year that the cars will reach 15,000.

The " set up " should look like the following quadratic equation -

20t² + 135t + 3050 = 15,000 - Isolate 0 on one side,

20t² + 135t - 11950 = 0 - From here on let us solve using the quadratic equation formula,

$t=\frac{-135+\sqrt{135^2-4\cdot \:20\left(-11950\right)}}{2\cdot \:20}:\quad \frac{-27+\sqrt{38969}}{8}$,

$t=\frac{-135-\sqrt{135^2-4\cdot \:20\left(-11950\right)}}{2\cdot \:20}:\quad -\frac{27+\sqrt{38969}}{8}$ ... now as you can see we have two solutions, but time can't be negative, and hence our solution is the first one - about 21.3 years. 1998 + 21.3 = ( About ) The year 2019. Therefore, in the year 2019 the number of new cars purchased will reach 15,000.