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VARVARA [1.3K]
3 years ago
12

Se deposita 4000 soles a una tasa de interés del 0,8% quincenas .¿Que interes producirá en cinco quincenas?

Mathematics
1 answer:
katen-ka-za [31]3 years ago
5 0

Answer:

Interes= 162,58

Step-by-step explanation:

Dada la siguiente información:

Se deposita 4000 soles a una tasa de interés del 0,8% quincenas

<u>Primero, debemos calcular el valor final de la inversión:</u>

VF= [VP*( 1 + i)^n]

VF= 4.000*(1,008^5)

VF= 4.162,58

<u>Para calcular el interés ganado en 5 quincenas, debemos usar la siguiente formula:</u>

Interes= 4.162,58  - 4.000

Interes= 162,58

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Answer:

48 pieces are cut.

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1/12 of an inch means that one inch would be 12 pieces. Since there are 4 pieces it would be 12x4= 48 pieces.

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Answer:

b.  \displaystyle \frac{1}{2}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}<u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                       \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle H(x) = \sqrt[3]{F(x)}<em />

<em />

<u>Step 2: Differentiate</u>

  1. Rewrite function [Exponential Rule - Root Rewrite]:                                      \displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}
  2. Chain Rule:                                                                                                        \displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]
  3. Basic Power Rule:                                                                                             \displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)
  4. Simplify:                                                                                                             \displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}
  5. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em> [Derivative]:                                                                              \displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}
  2. Substitute in function values:                                                                          \displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}
  3. Exponents:                                                                                                        \displaystyle H'(5) = \frac{6}{3(4)}
  4. Multiply:                                                                                                             \displaystyle H'(5) = \frac{6}{12}
  5. Simplify:                                                                                                             \displaystyle H'(5) = \frac{1}{2}

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

Unit: Derivatives

Book: College Calculus 10e

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Now we can plug that into the second equation to get y
y = x - 1, and we know that x = 4, so y = 4 - 1, y = 3. The solution is (4, 3)
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