using logarithms, find how long we left the money in the bank ( find t). with the bank returning $25,000 to us
487839
Need to know how much you invested at what interest rate first.
139495
Let's define some variables for the equation.\n\nD = monthly deposit ($200)\nr = annual interest rate (10% is 0.1)\nn = compound frequency (12 months in a year)\nA = account balance\ni = compounded interest\n\nLet's start the account this month, or month 0.\n\nA(0) = D\n\nOne month later, you make another deposit and earn some interest. But you only earn 1/12 of the 10% with you existing balance, which is r/12.  \n\nA(1) = D + A(0) + i(1)\n\nwhere, i(1) = A(0) * r/n. Which makes\n\nA(1) = D + A(0) + A(0) * r/n , or\nA(1) = D + A(0) * (1 + r/n ), or\nA(1) = D + A(0)*R, if R = 1 + r/n, but A(0) is D, so\nA(1) = D + D*R\n\nTwo months later, you make another deposit and earn some interest on the balance A(1). Similarly calculating,\n\nA(2) = D + A(1) * R, or\nA(2) = D + (D + D*R) * R, or\nA(2) = D + D*R + D*R^2\n\nKeep on going to three months later, \n\nA(3) = D + A(2) * R, or\nA(3) = D + D*R + D*R^2 + D*R^3, or\nA(3) = D * (1 + R + R^2 + R^3)\n\nFollowing this pattern, k months later, we get\n\nA(k) = D * (1 + R + R^2 + ... + R^k)\n\nNow, let's calculate the general expression (1 + R + R^2 + ... + R^k) in a closed form.\n\nLet S = (1 + R + R^2 + ... + R^k). Multiply R on both sides, we get\nS*R = (1 + R + R^2 + ... + R^k) * R, or\nS*R = R + R^2 + ... + R^k + R^(k + 1)\n\nSubtracting the two equations S*R and S on both sides, we get\n\nS*R - S = R + R^2 + ... + R^k + R^(k + 1)  - (1 + R + R^2 + ... + R^k)\n\nNotice the terms R all the way through R^k on the right hand side are cancelled and only left with R^(k + 1) - 1. So we have,\n\nS*R - S = S(R - 1) = R^(k + 1) - 1\n\nDividing (R - 1) on both side, we get\n\nS = [R^(k + 1) - 1] / (R - 1)\n\nRewriting the account balance at month k,\n\nA(k) = D * (1 + R + R^2 + ... + R^k), or\nA(k) = D*S = D * [R^(k + 1) - 1] / (R - 1)\n\nNow, 26 years make 26 * 12 = 312 months. But since you start with 0, so your k is actually 311. R = 1 + r/n = 1 + 0.1/12, and D = 200. So\n\nA(311) = 200 * [(1 + 0.1/12)^(311 + 1) - 1] / (1 + 0.1/12 - 1), or\nA(311) = 295667.15330119169 = 295667.15 (round to nearest penny)
506633
Easy. When the manager took $5 out of the $30 there was $25 remaining, right. That's $8.333333 paid each. When the bellboy gave them the $3 back that made it $9.333333 each. The remaining change from the $10 which is 67cents times 3 equals the $2 in the bellboys pocket.
156823
you collect 30.00 everyone takes1.00 back  =27.00\nrent is 25.00 ( that leaves 5.00 extra - 3.00 you already took back =  2.00 over
796583
Pretend that she invested x dollars at 6.5% and y dollars at 7%. You know that x+y=5000 (because she invested $5000 total), and .065x+.07y=340 (the amount of interest from both). From there, you are solving a system.\n\nWith a little help from my graphing calculator, I find that x=2000, y=3000.\n\nLet's check. $2000 dollars at 6.5% would give $130, and $3000 at 7% will give $210. $2000+$3000=$5000, and $130+$210=$340.
371728
straight line = 150,000 (200,000 * 1/12 *3 years)\ndouble-declining = 115,740.75 (1/12 * 200% = 16.66%)\n\n200,000*16.66% = 33,333.33\n166,666*16.66% = 27,777.77\n138,888*16.66% = 23,148.15\n\nAccumulated depreciation after 3 years = 84259.25\n200,000-84259.25=115,740.75\n\nAny more homework you need done, ha!
880662
I've got the first part.  I/rt = P\nSo, 93.08/(.065 times 1.33) = 1076.69 dollars\n.\nHere is the second one, I hope. I = Prt .\nSo, 56000times .12 times 1 = 6720\nNow divide 6720 by twelve to find the interest for each month.\nIt's 560 \nSubtract 560 from the monthly payment of 619.36 = 59.36. This is the amount coming off the principal. So after one month the principal is 56000- 59.36 = 55940.64\nHere is the third one.\nI = Prt\nSo 387000 times >147 times 15 = 853335 dollars to be paid back in 15 years. Divide that amount by 180 months(15 years) and you get a monthly payment of 4740.75\nI hope those folks have really good jobs. Ha
255187
sn777 is correct\n750 x .0125= 9.375 (for one month)\n9.375 x 6= 56.25 (for 6 months interest)\nadd your principle to interest:\n750+56.25 =  806.25 total due\n\nGen, $4 in interest total?
646947
1. A & B cross  -  2\n2. A returns    -  1\n3. C & D cross  -  10\n4. B returns    -  2\n5. A & B cross  -  2\n\nTotal minutes: 17\n:)
297329
one formula you can use is\n\nF = P(1 + i)^n\nwhere,\n\nF = future worth of the money\nP = present worth of the money\ni = interest rate\nn = number of years\n\nsince the future worth of the money will be twice of the present worth, then we can say that,\n\nF = 2P\n\nusing this in the formula you'll come up with,\n\n-> 2P = P(1 + i)^n  \n(divide the whole equation by P)\n-> 2 = (1 + i)^n \n(n is 20 years)\n-> 2 = (1 + i)^20\n(raising the whole equation by 1/20)\n-> 2^(1/20) = 1 + i\n-> 1.035264 = 1 + i\n-> i = 0.035264\n-> i = 3.5264%
45165
You could always look back in your checkbook to see how much you wrote the check for.  Best to call the old brokerage firm though and see if they still have the info in their databases.
303829
Just recalculate it each day and compound it again with the new balance that got taken over from yesterday.  Banks typically use 360 days you can use that when compounding.  If the balance is really changing daily, you can probably contact the bank once in a while to sync and make sure you're calculating them correctly.
1051751
Use yor calculator....
539778
If you're looking to find the present value of a single sum, that has accrued interest at a specific rate, the equation is:\n\nPV=amount / (1+i)^n\n\n\nEx: An amount of $350,000 with 12% interest for five periods(or years)\n\nPV=350,000 / (1+.12)^5\nPV=$198,599.39\nAn amount today of $350,000 that had been growing for 5 years at 12%, was $198,599.39 five years ago.
441515
times 2000 by %6.125 then divide by 2, thats added to the $2000, then multiply the new cash value by %6.125 and divide by 2, then repeat the process that i have gone through twice 2 more times\n\n%6.125 equals 0.06125
393533
I don't feel like working it out, but you just discount back the cash flows assuming they're constant, and then do some simple algebra.\n\nIt looks kind of like\n\n(785000 - 137) = (x / 1.07) + (x / 1.07^2) + ... (x / 1.07 ^10)\n\nSolve for x.
1095736
T=.00139x^2+.0304x=1.25 \n\nmaybe you meant\n\nT=.00139x^2+.0304x-1.25 \nor\nT=.00139x^2+.0304x+1.25 \n\n?????