Question

PLEASE HELP!!!!!!!!!!!!!!!!!!!! Three years ago, Andy invested $7,000 in an account that earns 10% interest compounded annually. The equation y = 7,000(1.10)t describes the balance in the account, where t is time in years. Andy made no additional deposits and no withdrawals. How much is in Andy’s account at this time? Round to the nearest dollar.

Answer 1

Replace t with 3 and evaluate.

Answer 2

23100???

Answer 3

Isn't that t an exponent?

Answer 4

Your answer is way too big.

Answer 5

THE T behind (1.10) is

Answer 6

So here is your equation:
\[y=7000(1.10)^3\]

Answer 7

What is 1.10 raised to the third power?

Answer 8

where did you get the 3 from

Answer 9

Three years ago Andy invested....
How much is in Andy’s account at this time?

Answer 10

What do you think t is?

Answer 11

nvm

Answer 12

my notes had something different

Answer 13

Like what?

Answer 14

Like it tells you the words if they are in the word problem you have to use it like annually= one per year, bi-monthly=6 times per year You would use the number in your equation

Answer 15

The formula is:
\[A=A _{0}(1+\frac{r}{n})^{nt}\]

Answer 16

http://www.sascurriculumpathways.com/portal/Launch?id=5064

Answer 17

loook

Answer 18

A is the final amount after t years
A_0 is the beginning amount.
r is the interest rate
n is the number of compounding periods in 1 year

Answer 19

did u see the website thats my notes

Answer 20

Yes. But I can only see the first page because I cannot log in.

Answer 21

Did you see the equation I posted?

Answer 22

yes i did . You have to use a number 1 in the equation because it says the word annually right ?

Answer 23

In your problem the beginning amount is 7000
n = 1 compounding period each year
r = .10
t = 3
So your problem becomes:
\[A=7000(1+\frac{.10}{1})^{1(3)}\]

Answer 24

Which is the same as:
\[A=7000(1.10)^3\]

Answer 25

Which is:
\[A=7000(1.331)\]

Answer 26

Which is:
\[A=$9317\]

Answer 27

theres the same example ^^ with the word annually

Answer 28

did u see it ??

Answer 29

Yep. Exactly what I just posted.

Answer 30

Did you see what I posted?

Answer 31

Im going over it & i understand it thanks! I hope its right

Answer 32

It is right.

Answer 33

yw

Answer 34

The value of Mr. Cox’s car x years after its purchase is given by the function, V(x) = 20,000(0.85)x. Approximately, what was the value of Mr. Cox’s car 5 years after its purchase? Round to the nearest dollar.

Answer 35

How would u do this ?

Answer 36

I would make x the exponent and replace it with 5

Answer 37

(0.85)^x

Answer 38

so then you would just multiply it ?

Answer 39

\[V(5)=20000(.85)^5=20000(.4437)=8874.10\]

Answer 40

$8874 to the nearest dollar.

Answer 41

thats the answer ??

Answer 42

if it is it it rounded

Answer 43

yep

Answer 44

thanks

Answer 45

yw

Answer 46

can u help me on this one too ? A city’s current population is 1,000,000 people. It is growing at a rate of 3.5% per year. The equation P = 1,000,000(1.035)x models the city’s population growth where x is the number of years from the current year. In approximately how many years will the population be 1,200,000? Round to the nearest tenth.

Answer 47

Replace P with 1,200,000 and solve for x.

Answer 48

1,200,000=1,000,000(1.035)x ?

Answer 49

yes

Answer 50

would you multiply 1,000,000(1.035)x

Answer 51

opps its supposed to be (1.035)^x

Answer 52

I know.

Answer 53

.8625 ??

Answer 54

is this right

Answer 55

you there ???

Answer 56

I haven't done it yet.

Answer 57

\[1.2=1.035^x\]
\[\log_{10}1.2=x \log_{10}1.035 \]
\[\frac{\log_{10} 1.2}{\log_{10}1.035}=x \]
x=5.3 years

Answer 58

do i round ?

Answer 59

What does the problem say to do?

Answer 60

round to the nearest tenth so its 10 !!

Answer 61

I tried 1 and it was wrong .

Answer 62

or is it 5 ?

Answer 63

The nearest TENTH not the nearest 10. Do you know place value?

Answer 64

yes