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lutik1710 [3]
3 years ago
11

Blanche bought a 5-year cd for $7100 with an apr of 2.8%, compounded quarterly, but she wants to take all her money out 9 months

early. The early redemption fee for the cd is 3 months' interest on the original principal. What is the periodic interest rate of the cd
Mathematics
1 answer:
lesantik [10]3 years ago
8 0

Answer:

The money she will end up earning in interest on the cd = $11,352.90

Step-by-step explanation:

The formula for getting the accumulated amount(compounded) is;

A =P(1+\frac{r}{n})^n*t

Where

A = Accumulated amount  

P = principle (deposit)

r = interest rate and

n = no of times interest applied per time period.  

The interest is compounded quarterly so in one year it will be 4 times

In 5 years

n = (5×4)-3 = 17  (as she will withdraw 3 month before the completion of five years)

A = 7100(1+\frac{2.8}{100} )^17

  = 7100( 1 + 0.028)^17

  =  7100(1.028)^17  

   = 7100 * 1.599

  = 11,352.90

Therefore the money she will end up earning in interest on the cd = $11,352.90

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Answer:\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Trigonometric Identities

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Integration

  • Integrals
  • Definite/Indefinite Integrals
  • Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                    \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

U-Substitution

  • Trigonometric Substitution

Reduction Formula:                                                                                               \displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution (trigonometric substitution).</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle x = sin(u)
  2. [<em>u</em>] Differentiate [Trigonometric Differentiation]:                                         \displaystyle dx = cos(u) \ du
  3. Rewrite <em>u</em>:                                                                                                       \displaystyle u = arcsin(x)

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Trigonometric Substitution:                                                           \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du
  2. [Integrand] Rewrite:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du
  3. [Integrand] Simplify:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du
  4. [Integral] Reduction Formula:                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b
  5. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du
  6. [Integral] Reduction Formula:                                                                          \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  7. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  8. [Integral] Reverse Power Rule:                                                                     \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  9. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b
  10. Back-Substitute:                                                                                               \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b
  11. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  12. Rewrite:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  13. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:              \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

8 0
3 years ago
Read 2 more answers
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