Hi there
The formula is
A=p (1+r/k)^kt
A future value 3000
P present value 100
R interest rate 0.02
K compounded monthly 12
T time?
We need to solve for t
T=[log (A/p)÷log (1+r/k)]÷k
T=(log(3,000÷100)÷log(1+0.02÷12))÷12
T=170.202 years
So it's a
Hope it helps
Well, the house is increasing by 10% each year, so whatever the current price is, the new price is THAT plus 10%, well, since THAT is 100%, shouldn't that be 100% + 10%, or 110%?
![\bf \qquad \textit{Amount for Exponential Growth}\\\\ A=I(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &120,000\\ r=rate\to 10\%\to \frac{10}{100}\to &0.10\\ t=\textit{elapsed time}\\ \end{cases} \\\\\\ A=120000(1+0.10)^t\implies \begin{array}{lclll} A=120000(&1.1&)^t\\ &\uparrow \\ &rate\\ &of\\ &growth \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BAmount%20for%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DI%281%20%2B%20r%29%5Et%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AI%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%26120%2C000%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.10%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D120000%281%2B0.10%29%5Et%5Cimplies%20%0A%5Cbegin%7Barray%7D%7Blclll%7D%0AA%3D120000%28%261.1%26%29%5Et%5C%5C%0A%26%5Cuparrow%20%5C%5C%0A%26rate%5C%5C%0A%26of%5C%5C%0A%26growth%0A%5Cend%7Barray%7D)
1.1 is the decimal format, but you can simply multiply it by 100 to get the percentage, 1.1 * 100 = 110%
Answer:
5/4 is the correct answer for this question
Answer:
u^8/8 - 4u^7/7 - u^5/5 + 2u + C
Step-by-step explanation:
Find the solution to the integral question in the attachment.
Given ∫(u7 − 4u6 − u4 + 6/3)du
The solution to the integral is expressed as;
u^8/8 - 4u^7/7 - u^5/5 + 2u + C