Question

The value V(t) of a $18,000 car after t years that depreciates at 20% per year is given by the formula V(t) = V0(b)t. What is the value of V0 and b, and what is the car's value after 5 years (rounded to the nearest cent)?

Hint: Use the equation b = 1 − r to determine depreciation.

V0 = $18,000, b = 0.80, and the value after 5 years is $5,898.24
V0 = $18,000, b = 0.20, and the value after 5 years is $5.76
V0 = $18,000, b = 1.20, and the value after 5 years is $44,789.76
V0 = $18,000, b = 0.80, and the value after 5 years is $5,760.00

Answer 1

@GreenCat are you able to help me with this?

Answer 2

no sorry

Answer 3

Hint: Use the equation b = 1 − r

that is not much of a hint, unless you know what "r" is supposed to be.
Do you know what they mean by r ?

Answer 4

It means doubling or 2 in this case. @phi

Answer 5

it can't mean that.
they say depreciates at 20% per year
depreciate means "go down in value"

I guess they mean r is 0.2 (20% written as a decimal)
so "b" in your formula , using the hint is 1-0.2

Answer 6

Ok that makes sense. This lesson has been really hard for me to understand. Is there anyway that you might be able to talk me through this question? @phi

Answer 7

If you have time

Answer 8

Yes I do. I'm pretty sure that it's either a or d. Because b= 1-0.2 = 0.8

Answer 9

yes.

Answer 10

First step

Do you know what 10% of 100 is ?
i.e. do you know how to find 10% of 100 ?

Answer 11

It would be 10/100, which is 10

Answer 12

you would write 10% as 10/100 (which simplifies to 0.1 as a decimal)
then you do 0.1 * 100
you get 10

can you do 10% of 50 ?

Answer 13

It's 5

Answer 14

yes. But I want to make sure you get the idea
say you want 1/2 of 20
you would multiply 1/2 * 20 and get 10

or 1/3 of 90
you would multiply 1/3* 90 and get 30
does that make sense?

Answer 15

percents are the same idea, except the fraction always has 100 in the bottom
20% is the fraction 20/100

and 20% of 100 you do by multiplying the fraction 20/100 * 100
you would get 20

Answer 16

Yes that makes sense. I'm following you

Answer 17

That is the first idea that you have to be comfortable with.

Now the second idea

100 is reduced by 10%

that is short-hand (i.e. they leave some words out)
for
100 is reduced by 10% of 100
and that means
100 minus 10% of 100

or in math
100 - 0.1*100
do you know what that is ?

Answer 18

Would that be 9990?

Answer 19

first what is
0.1*100 ?

Answer 20

It's 99.9

Answer 21

you calculator is broken
what is
\[ \frac{100}{10} \]

Answer 22

I'm sorry I do not know how I got that. It's 10. 0.1*100 = 10

Answer 23

yes. If you are not good with arithmetic, make a note to study it more. (Khan has lots of videos )

In the meantime, you can survive using a calculator

Answer 24

back to the 2nd idea

something costs $20
then they say, "on sale! 10% off"

when they say 10% off they mean
1) find 10% of 20
can you do that ?

Answer 25

10/100 *20 =2

Answer 26

yes,
something costs $20
then they say, "on sale! 10% off"

when they say 10% off they mean
1) find 10% of 20, which is 2
2) take 2 off of 20, i.e. figure out 20 -2

Answer 27

in other words, the new price will be $18

ok so far?

Answer 28

the next idea is
is to write 10% off as
\[ 20- \frac{10}{100} 20 = 20\left(1-\frac{10}{100} \right) \]
that uses the "distributive property"
1 - 0.1 = 0.9

Answer 29

in other words,
10% of 20 can be written as
20 - 20*0.1 = 20(1-0.1) = 20*0.9

if you multiply 20*0.9 you get 18 (the same answer as before)

Answer 30

Yes! My computer messed up for a sec. I think I actually figured out the answer. If you do the equation this way. 18,000(0.80)^5 you get 147,456/25. And if you simplify that, you get 5898.24. And that is choice A!!!

Answer 31

Is that right?

Answer 32

I'm not sure how you got 147456/25, but yes that is the correct answer
on my calculator I would do 0.8^5 and get 0.32768
then multiply by 18000

Answer 33

I have a few more that I need help with. Do you have time to help me?

Answer 34

one more

Answer 35

The half-life of a substance is how long it takes for half of the substance to decay or become harmless (for certain radioactive materials). The half-life of a substance is 8.6 days and there is an amount equal to 15 grams now. What is the expression for the amount A(t) that remains after t days, and what is the amount of the substance remaining (rounded to the nearest tenth) after 37 days?

Hint: The exponential equation for half-life is A(t) = A0(0.5)t/H, where A(t) is the final amount remaining, A0 is the initial amount, t is time, and H is the half-life.

A(t) = 15(0.5)8.6t, 0.0 gram remaining
A(t) = 15(0.5)t/8.6, 0.8 gram remaining
A(t) = 8.6(15)(0.5)t, 0.0 gram remaining
A(t) = 15(0.5)8.6/t, 12.8 grams remaining

Answer 36

Hint: The exponential equation for half-life is A(t) = A0(0.5)t/H, where A(t) is the final amount remaining, A0 is the initial amount, t is time, and H is the half-life

to use their formula you have to interpret the info they gave you.
For example, you need to know A0 , which is the "initial amount"
any idea what the initial amount is ?

Answer 37

I think that it is 15. Is that right?

Answer 38

yes. they say

there is an amount equal to 15 grams now.

so you start with 15 grams (initial means starting or first. Like initial is the first letter of your name)

Answer 39

so put in 15 into the equation
\[ A(t) = A0(0.5)^{\frac{t}{H}} \\ A(t) = 15(0.5)^{\frac{t}{H}}
\]

Answer 40

you also need
H is the half-life

what is H ?

Answer 41

the half life would be 8.6?

Answer 42

so we put that in
\[ A(t) = 15(0.5)^{\frac{t}{8.6}} \]

Answer 43

then 37 would be t?

Answer 44

they say "t" is time (starting from now) in days
so put in 37 for t
\[ A(t) = 15(0.5)^{\frac{37}{8.6}} \]

Answer 45

I guess I should write A(37) =

Answer 46

I got 0.7602601167 and then when I round to the nearest tenth it would be 0.8. Right?

Answer 47

yes. But only one of the choices has the correct formula, so we did not need to find the actual number

Answer 48

So the correct answer would be B?

Answer 49

yes

Answer 50

Thank you so much! It makes so much more sense! I really appreciate it. I only have one more question now. You really helped me :) Thank you