In 2003, the price of a certain automobile was approximately $32,000 with a depreciation of $1,740 per year. After how many years will the car's value be $23,300? a) Write an equation to model the problem. Let t represent the number of years after 2003. For example, the year 2005 would be represented by t = 2.

The correct answer is 5 years i.e. 2008.

Step-by-step explanation:

Price of the automobile in 2003 is $32000.

Depreciation per year is given by $1740.

Therefore let the car value is depreciated for t number of years.

Value depreciated for t years is given by $ (1740t).

The final value of the car after t years is given to be $23300.

Thus the equation is given by 32000 - 1740t = 23300.

⇒ 1740t = 32000 - 23300

⇒ 1740t = 8700

⇒ t = 5

Thus after 5 years the value of the car is $23300.

Thus in 2008 the depreciated price of the car would be $23300.

p

step-by-step explanation:

the answer is c.) yes, [triangle]klp can be reflected across the line containing [line]kp and then translated so that p is mapped to m.

the reason is this:

looking at the triangles visually we can see they share the same features. using rigid transfromations we can reflect it over [line]kp then all we need to do is translate up.