Sense we know the values of a and b all we need to do is fill them in the problem and solve it So once you fill them in it would give you -4+(4•-3) Then you would simplify the parentheses -4+(-12) After wards you would add these two numbers together and since they are negative it would appear for the number to be bigger even though it is getter quite smaller So the answer to this expression would be = -16
So is doubling, first off, meaning, if the current amount say P, then when it doubles is 2P, or double that, or P + P, so the rate of growth is 100%, since it's doubling.
![\bf \textit{Periodic/Cyclical Exponential Growth}\\\\ A=P(1 + r)^{\frac{t}{c}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\to &650\\ r=rate\to 100\%\to \frac{100}{100}\to &1.00\\ t=\textit{elapsed time}\to &10\\ c=period\to &9 \end{cases} \\\\\\ A=650(1 + 1)^{\frac{10}{9}}\implies A=650(2)^{\frac{10}{9}}\implies A=650\sqrt[9]{2^{10}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BPeriodic%2FCyclical%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DP%281%20%2B%20r%29%5E%7B%5Cfrac%7Bt%7D%7Bc%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%26650%5C%5C%0Ar%3Drate%5Cto%20100%5C%25%5Cto%20%5Cfrac%7B100%7D%7B100%7D%5Cto%20%261.00%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%2610%5C%5C%0Ac%3Dperiod%5Cto%20%269%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D650%281%20%2B%201%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%282%29%5E%7B%5Cfrac%7B10%7D%7B9%7D%7D%5Cimplies%20A%3D650%5Csqrt%5B9%5D%7B2%5E%7B10%7D%7D)