The value of a car decreases $500 for every 1,000 miles it is driven. the current value of the car is $23,000, and the car is driven an average of 10,000 miles per year. what will be the value of the car, in dollars, at a point in time y years from now? a) $23,000 − $500y b) $23,000 − $0.02y c) $23,000 − $5,000y d) $23,000 − $0.0002y

A) $23,000 − $500y

Step-by-step explanation:

Given,

The current value of the car = $ 23,000

∵ Car is driven an average of 10,000 miles per year.

So, the number of miles it has been driven in y years = 10,000y

Also, the value of a car decreases $500 for every 1,000 miles it is driven,

i.e., for 1000 miles, decrement in price = $ 500

⇒ For 1000y miles, decrement in price = 500y dollars,

⇒ For y years, decrement in price = 500y dollars

Hence, the final value of car after y years = current value of car - total derment in price

= $ 23000 - $ 500y

Option 'A' is correct.

D

Step-by-step explanation:

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b