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Suppose the amount invested (or borrowed) is K, Suppose the rate of interest is R% annually, Suppose the amount accrues interest for Y years. Then the interest I is 100*K[(1 + R/100)^Y - 1]
just add all the numbers I suppose and the number you get is the number you graph the solution. I suppose.
The formula for the amount received (A) when investing P after n periods where the rate per period is r% is given by:A = P(1+ r/100)nIf you have an apr (annual percentage rate) and it is applied monthly, then the rate is apr ÷ 12 applied 12n times (for n years).So going back to your question:I assume the 4% is 4% apr.So the return if the interest is applied yearly would be:A = 100 x (1 + 0.04)7~= 131.59However, you specify continuously - what exactly does "continuously" mean. Let suppose that the interest is applied monthly, then:A = 100 x (1 + 0.04/12)7 x 12~= 132.25Hmmm...we got a little more; how about more often, say daily (I'll ignore the fact of leap years and assume 365 days per year):A = 100x (1 + 0.04/365)7 x 365~= 132.31Not much more, how about hourly (still assuming 365 days/year):A = 100x (1 + 0.04/8760)7 x 8760~= 132.31(There's actually a difference, but not big enough to show in monetary terms). Going to every minute:A = 100x (1 + 0.04/525600)7 x 525600~= 132.31We seem to have hit a limit!The actual return you will get if the (100r)% apr was applied continuously for n years is:A = PernSo for 100 at 4% apr for 7 years, this is:A = 100e0.04 x 7~= £132.31
13% daily is worse than any loan shark!Suppose it takes n days, then38500 = 19000*(1 + 13/100)n= 19000*(1.13)nSo 38500/19000 = 1.13nln(38500/19000) = n*ln(1.13)so that n = ln(38500/19000) / ln(1.13) = 5.78So 6 days.
The answer depends on whether you are dealing with simple interest of compound interest. Suppose P = Principle R = Rate (in % per annum) T = Time (in years) I = Interest Then for simple interest: I = P*R*T/100 so that P = 100*I/(R*T) For compound interest P+I = P*(1+R/100)T so that P = I/[(1+R/100)T - 1]
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390.45
SupposeCapital invested = YAnnual Interest Rate = R%Period of investment = TThen if the interest is calculated (and compounded) n times a yeartotal value =Y*[1 + r/(100*n)]^(n*T)So interest accrued = Total value - YSupposeCapital invested = YAnnual Interest Rate = R%Period of investment = TThen if the interest is calculated (and compounded) n times a yeartotal value =Y*[1 + r/(100*n)]^(n*T)So interest accrued = Total value - YSupposeCapital invested = YAnnual Interest Rate = R%Period of investment = TThen if the interest is calculated (and compounded) n times a yeartotal value =Y*[1 + r/(100*n)]^(n*T)So interest accrued = Total value - YSupposeCapital invested = YAnnual Interest Rate = R%Period of investment = TThen if the interest is calculated (and compounded) n times a yeartotal value =Y*[1 + r/(100*n)]^(n*T)So interest accrued = Total value - Y
Suppose the amount invested (or borrowed) is K, Suppose the rate of interest is R% annually, Suppose the amount accrues interest for Y years. Then the interest I is 100*K[(1 + R/100)^Y - 1]
just add all the numbers I suppose and the number you get is the number you graph the solution. I suppose.
I suppose that is not possible.
I suppose that you think to the dissolution of salt in water.
To find the grams of sodium fluoride needed, use the formula: grams = moles x molar mass. First calculate the moles by multiplying the volume (6.3 L) by the molarity (3.6 mol/L). Then multiply the moles by the molar mass of sodium fluoride (41.99 g/mol) to find the grams required. In this case, approximately 920 grams of sodium fluoride are needed.
I don't know, I suppose we have to ask a chemist.
I suppose that you think to a buffer, not bond.
No. I is as described for the stated period.
At a certain temperature, Steel becomes liquid. And I suppose it is even a solution as Steel is an alloy of metals, Not only Iron alone.