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I have no examples. So please bare with me.
A new car worth $24,000 is depreciating in value by $3,000 per year. Complete A- B. A. Write the formula that models the cars value, y, in dollars, after x years.
This is question 5.
A. If the car's value declines by 3000/year, then this can be represented by 3000x, and because the car originally cost 24000, the equation would be y=24000-3000x
B. Use the formula from part A to determine after how many years the cars value will be $15,000.
Here, you would just plug in 15000 for y. 15000=24000-3000x There are a number of ways you can go about solving this, but in the end x has to be isolated on one side of the equation. Personally, I like to subtract and then divide. First, I would subtract 24000 from both sides -9000=-3000x And then I divide by -3000. 3=x So three years is your answer.
Okay. Thank you. Would you mind helping with a few more?
do you know if we should assume that a month is 30 days?
I honestly don't know
@Vocaloid what do you think?
The help me solve this button says " Here, x is the number of times in a month the bus is used."
Maybe its 1.75x=30-0.75x? That would make the most sense to me.
Okay.. I have new numbers now since I clicked help me solve this... 1.50x= 24x- .50x
= 24 not 24x
I got x=12