John, Sally, and Natalie would all like to save some money. John decides that it would be best to save money in a jar in his closet every single month. He decides to start with $300, and then save $100 each month. Sally has $6000 and decides to put her money in the bank in an account that has a 7% interest rate that is compounded annually. Natalie has $5000 and decides to put her money in the bank in an account that has a 10% interest rate that is compounded continuously.
What type of equation models John’s situation? ______
Write the model equation for John’s situation________
How much money will John have after 2 years? _______
How much money will John have after 10 years? ________
dude just wants the entire answer i'm sure
girth.
The whole thing. I don't understand it.
You want to multiply the vector scalar projected onto the dot product of the terminal crosss product's velocity. Basically, the answer is relative to the proximity to gargantua's event horizon...
-Kip Murz Thorne
To model John's situation, we can use a linear equation. A linear equation has the general form: y = mx + b, where y represents the amount of money, x represents the number of months, m represents the rate of change (saving $100 each month), and b represents the initial amount ($300).
The model equation for John's situation can be written as: y = 100x + 300.
To find out how much money John will have after a certain number of years, we need to convert the years into months. Since there are 12 months in a year, we multiply the number of years by 12 to get the total number of months.
After 2 years:
x = 2 years * 12 months/year = 24 months
Now we substitute x = 24 into the equation:
y = 100 * 24 + 300
y = 2400 + 300
y = 2700
John will have $2700 after 2 years.
After 10 years:
x = 10 years * 12 months/year = 120 months
Now we substitute x = 120 into the equation:
y = 100 * 120 + 300
y = 12000 + 300
y = 12300
John will have $12300 after 10 years.
Where are you stuck on this?
We have the following growth functions
John: 300 + 100x after x months
Sally: 6000 * 1.07^x after x years
now, what do you know about continuous interest?