Math 22 Natural Exponential Functions - Revision history

Tphan046: /* Present Value */

2020-08-11T15:12:24Z

Present Value

2020-08-11T15:08:27Z

2020-08-11T15:06:32Z

Compound Interest

2020-08-11T15:04:03Z

Compound Interest

2020-08-11T15:03:49Z

Compound Interest

2020-08-11T15:03:30Z

Compound Interest

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2020-08-11T15:00:57Z

2020-08-11T14:57:40Z

2020-08-11T14:57:26Z

@@ Line 50: / Line 50: @@
 <math>P=\frac{A}{(1+\frac{r}{n})^{n t}}</math>
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 47: / Line 47: @@
 |}
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 7: / Line 7: @@
    Let <math>P</math> be the amount deposited, <math>t</math> the number of years, <math>A</math> the balance,
    and <math>r</math> the annual interest rate (in decimal form).
-. Compounded <math>n</math> times per year: <math>A=P(1+\frac{r}{n})^{rt}</math>
+. Compounded <math>n</math> times per year: <math>A=P(1+\frac{r}{n})^{nt}</math>
 . Compounded continuously: <math>A=Pe^{rt}</math>
@@ Line 16: / Line 16: @@
 !Solution: &nbsp;
 |-
-|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+|<math>A=3000(1+\frac{0.04}{4})^{(4)10}</math>
 |}
-'''a)''' Quarterly
+'''a)''' Annually
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+|<math>A=3000(1+\frac{0.04}{1})^{(1)10}</math>
 |}
@@ Line 30: / Line 30: @@
 !Solution: &nbsp;
 |-
-|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+|<math>A=3000(1+\frac{0.04}{12})^{(12)10}</math>
 |}
@@ Line 37: / Line 37: @@
 !Solution: &nbsp;
 |-
-|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+|<math>A=3000(1+\frac{0.04}{365})^{(365)10}</math>
 |}

@@ Line 10: / Line 10: @@
 . Compounded continuously: <math>A=Pe^{rt}</math>
-'''Exercises''' Find the balance in an account when <math>\$3000</math> is deposited for 10 years at an interest rate of 4%, compounded as follows.
+'''Exercises''' Find the balance in an account when $3000 is deposited for 10 years at an interest rate of 4%, compounded as follows.
 '''a)''' Quarterly

@@ Line 10: / Line 10: @@
 . Compounded continuously: <math>A=Pe^{rt}</math>
-'''Exercises''' Find the balance in an account when <math>\$3000</math> is deposited for 10 years at an interest rate of 4, compounded as follows.
+'''Exercises''' Find the balance in an account when <math>\$3000</math> is deposited for 10 years at an interest rate of 4%, compounded as follows.
 '''a)''' Quarterly

@@ Line 44: / Line 44: @@
 !Solution: &nbsp;
 |-
-|<math>A=Pe^{rt}=3000(e^{0.04}{10})</math>
+|<math>A=Pe^{rt}=3000(e^{(0.04) (10)})</math>
 |}

@@ Line 3: / Line 3: @@
    <math>\lim_{x\to 0} (1+x)^{\frac{1}{x}}=e</math>
    <math>e\approx 2.71828182846</math>
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''