Amount owed at the end of 1 year is 3640
<h3><u>Solution:</u></h3>
Given that yoko borrows $3500.
Rate of interest charged is 4% compounded each year
Need to determine amount owed at the end of 1 year.
In our case
:
Borrowed Amount that is principal P = $3500
Rate of interest r = 4%
Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1
<em><u>Formula for Amount of compounded yearly is as follows:</u></em>
Where "p" is the principal
"r" is the rate of interest
"n" is the number of years
Substituting the values in above formula we get
Hence amount owed at the end of 1 year is 3640
Answer:
<h2>5.5x+10<=200</h2>
Step-by-step explanation:
5.5x+10 less than or equal to 200
Yes. they may be very small or big but they both can be represented either way.
Answer:
She can create a function for the total savings in terms of the time x by adding the two functions together. The new function should look like this:
T(x)=S(x)+a(x)
So now we can substitute the given functions:
T(x)=450+6(x-2)
But this function can be simplified for it to be easier for her to calculate her total amount of savings. We can do that by distributing the 6 into the parenthesis. This is, multiply the 6 by both the x and the -2, so we get:
T(x)=450+6x-12
and now we can combine like terms. We can subtract 12 from 450 so we get:
T(x)=6x+438
And that will be our final function.