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shutvik [7]
3 years ago
6

Pls help asappppppp it is very important I need it pls

Mathematics
1 answer:
Volgvan3 years ago
5 0

Given:

The graph of a proportional relationship.

To find:

The constant of proportionality, the value of y when x is 24 and the value of x when y is 108.

Solution:

If y is directly proportional to x, then

y\propto x

y=kx             ...(i)

Where, k is the constant of proportionality.

The graph of proportional relationship passes through the point (5,15).

Substituting x=5 and y=15 in (i), we get

15=k(5)

\dfrac{15}{5}=k

3=k

Therefore, the constant of proportionality is 3.

Substituting k=3 in (i) to get the equation of the proportional relationship.

y=3x              ...(ii)

Substituting x=24 in (ii), we get

y=3(24)

y=72

Therefore, the value of y is 72 when x is 24.

Substituting y=108 in (ii), we get

108=3x

\dfrac{108}{3}=y

36=y

Therefore, the value of x is 36 when y is 108.

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Step-by-step explanation:

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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

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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
  2. [Integral] U-Substitution:                                                                              \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du
  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}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)
  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)

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

Unit: Integration

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Read more on Brainly.com - brainly.com/question/8678447#readmore

Step-by-step explanation:

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