Interest rate i=19.1% compounded annually.
Initial deposit, P=48900
Time t=53 months=53/12 years

Use compound interest formula
Future value,
F=P(1+i)^(t)=48900(1+0.191)^(53/12)
= $105,824.40 (to the nearest cent)

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%20%20%5Csqrt%7B3x%7D%20%7D%7B1%20%2B%20e%5E%7B2x%7D%20%7D%20" id="TexF

kumpel [21]

$~~~~~~~~y = \dfrac{\sqrt{3x}}{ 1+ e^{2x}}\\\\\\\implies \dfrac{dy}{dx} = \dfrac{d}{dx} \left( \dfrac{\sqrt{3x}}{1+ e^{2x}} \right)\\\\\\~~~~~~~~~~~~=\dfrac{(1 + e^{2x} ) \dfrac{d}{dx} \left(\sqrt{3x} \right) - \left(\sqrt{3x} \right)\dfrac{d}{dx}(1+e^{2x})}{\left( 1+ e^{2x} \right)^2}~~~~~~~~~~~~;[\text{Quotient rule}]\\\\\\~~~~~~~~~~~~=\dfrac{(1+e^{2x}) \cdot \dfrac{1}{2\sqrt{3x}} \cdot 3-\left(\sqrt{3x} \right) \cdot 2 e^{2x}}{(1 + e^{2x})^2}~~~~~~~~~~~~~~~~~~~~;[\text{Chain rule}]\\\\\\$

$=\dfrac{\dfrac{\sqrt 3 (1 + e^{2x})}{2\sqrt x} -2e^{2x} \sqrt{3x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\tfrac{\sqrt 3(1 +e^{2x}) - 4x\sqrt 3 e^{2x}}{2\sqrt x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\sqrt 3(1+e^{2x} -4xe^{2x})}{2\sqrt x(1 +e^{2x})^2}$

7 0

1 year ago