A person borrows $10,000 and repays the loan at the rate of $2, 400 per year. The lender charges interest of 10% per year. Assuming the payments are made continuously and interest is compounded continuously (a pretty good approximation to reality for long-term loans), the amount M(t) of money (in dollars) owed t years after the loan is mace satisfies the differential equation
dM/dt = 1/10 M - 2400 and the initial condition M(0) = 10000.
(a) Solve this initial-value problem for M(t).
M(t) = -1400e^(r/10) + 2400
(b) How long does it take to pay off the loan? That is, at what time t is M(t) = 0? Give your answer (in years) in decimal form with at least 3 decimal digits.
3.80211 x years

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

the answer is -6

step-by-step explanation:

the answer is -6 because its next to the variable m