NEED ANSWERS ASAP In the year 2005, the average cost of a car could be modeled by the equation C= -15x2

5 answers
Question:

NEED ANSWERS ASAP In the year 2005, the average cost of a car could be modeled by the equation C= -15x2 + 20x - 3 where x is the number of years since 2005. By the year 2010 the average cost had changed, and the equation could be modeled by C= -10x2 + 30x - 2. Find the difference in average cost equation for cars between 2005 and 2010. In the following graph of F(x) = -(2/3)(x-3)2 + 2 is the preimage of a transformation of G(x)which is the image. What is the mapping statement for the function G(x)?

Answers

60 dollars.

Step-by-step explanation:

I got this by plugging in the number 5 for x since it said that x equals the amount of years it has been since 2005. I got 5 since it is talking about the relationship between 2005 and 2010 (2010-2005=5). Therefore you would get the equation -15 x 2 + 20 (5) -3 = -10 x 2 + 30 (5) -3. First you would do the left side of the equation, start with multiplication from left to right and then addition and subtraction from left to right. You would then get 67 = -20 + 30 (5) -3. You would again start with the multiplication and the addition and subtraction from left to right. You would end up with 67 equals 127. You would subtract 167 from 127 to get 60 dollars which is the difference from the years. I hope this helps and is right.

ANSWER

[tex]C_{2010} - C_{2005} = 5 {x}^{2}+ 10x + 1[/tex]

EXPLANATION

The average cost equation in the year 2005 is:

[tex]C_{2005}</p<p= - 15 {x}^{2} + 20x - 3[/tex]

The average cost equation in the year 2010 is

[tex]C_{2010}</p<p= - 10 {x}^{2} + 30x - 2[/tex]

The difference in average cost equation for cars between 2005 and 2010 is

[tex]C_{2010} - C_{2005} = - 10 {x}^{2} + 30x - 2 - ( - 15 {x}^{2} + 20x - 3)[/tex]

[tex]C_{2010} - C_{2005} = - 10 {x}^{2} + 30x - 2 + 15 {x}^{2} - 20x + 3[/tex]

We regroup the terms to get,

[tex]C_{2010} - C_{2005} = - 10 {x}^{2} + 15 {x}^{2}+ 30x - 20x + 3 - 2[/tex]

Combine the similar terms to

[tex]C_{2010} - C_{2005} = 5 {x}^{2}+ 10x + 1[/tex]

C_{2010} - C_{2005} = 5 {x}^{2}+ 10x  + 1

Step-by-step explanation:

The average cost equation in the year 2005 is:

The average cost equation in the year 2010 is

The difference in average cost equation for cars between 2005 and 2010 is

We regroup the terms to get,

Combine the similar terms to

Remark

The difference is C_2005 - C_2010

Calculation

C_2005 - C_2010 = (-15x^2 + 20x - 3) - (- 10X^2 + 30x - 2)     Remove the brackets

C_2005 - C_2010 = - 15x^2 +20x - 3 + 10^2 - 30x + 2 Look how the signs change in the second set of brackets.

C_2005 - C_2010 = -5x^2 - 10x - 1 Answer

 1

Explanation:

Plug x = 0 into the first equation to find that C = -3. We use x = 0 since 0 years have passed by (the starting point is 2005 for this equation).

Now plug x = 0 into the second equation. The starting point is now 2010 which explains why we use the same x value, just for a different equation. You should get C = -2 here.

The difference from C = -3 to C = -2 is 1, as this is the distance between the two values.

Chances are C is measured in thousands of dollars, so C = 1 represents an average cost of 1000 dollars. Though your teacher never mentions "in thousands of dollars", so it's probably best to stick to 1 instead of 1000. I would ask your teacher to clarify.

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