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vodomira [7]
2 years ago
11

100 points + Brainliest Please help! Please give steps!!

Mathematics
1 answer:
Mrac [35]2 years ago
6 0

Answer:

\textsf{A)}\quad \dfrac{5}{10}, \:\: -\dfrac{6}{11}, \:\: \dfrac{7}{12}, \:\: -\dfrac{8}{13}

\textsf{B)} \quad \text{f}\:\!(n)=\dfrac{n}{n+5}    

If n is odd then f(n) is positive.

If n is even then f(n) is negative.

C)  positive

Step-by-step explanation:

Given sequence:

\dfrac{1}{6},\:\: -\dfrac{2}{7},\:\: \dfrac{3}{8}, \:\:-\dfrac{4}{9},...

<h3><u>Part A</u></h3>

From inspection of the sequence the pattern is:

  • Add 1 to the numerator of the previous term.
  • Add 1 to the denominator of the previous term.
  • Make every other term negative.

Therefore, assuming the pattern continues, the next four terms in the sequence will be:

\dfrac{5}{10}, \:\: -\dfrac{6}{11}, \:\: \dfrac{7}{12}, \:\: -\dfrac{8}{13}

<h3><u>Part B</u></h3>

An <u>explicit formula</u> for an arithmetic sequence allows you to find the nth term of the sequence.

\textsf{First term}: \quad \text{f}(1)=\dfrac{1}{6}

\textsf{Second term}: \quad \text{f}(2)=-\dfrac{2}{7}

From inspection of the sequence, we can see that the <u>numerator</u> is the position of the term (n) in the sequence.  So when n = 1 the numerator is 1, when n = 2 the numerator is 2, etc.

The denominator is 5 more than the position of the term in the sequence.  So when n = 1 the denominator is 1 + 5 = 6, and when n = 2 the denominator is 2 + 5 = 7.

We can also observe that if the numerator (the value of n) is odd then the term is positive, and if the numerator  (the value of n) is even then the term is negative.

Therefore, the rule for the nth term is:

\text{f}\:\!(n)=\dfrac{n}{n+5}

If n is odd then f(n) is positive.

If n is even then f(n) is negative.

<h3><u>Part C</u></h3>

53 is an odd number.

As the function f(n) is positive when n is odd, the sign of f(53) is positive.

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Answer:

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Basic Power Rule:

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Integration

  • Integrals

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

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx

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

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [Bounds] Switch:                                                                                           \displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.

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

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx
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  3. [Integral] Exponential Integration:                                                               \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}
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  5. Simplify:                                                                                                         \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

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Step-by-step explanation:

The given expression is "three less than the quotient of ten and a number, increased by six”

The quotient of ten and a number increased by six is translated into mathematical expression as:

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Three less than the quotient of ten and a number, increased by six is

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When we substitute n=2, we get:

(\frac{10}{2}+6)-3

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Answer: The flowchart is shown below.

Explanation:

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