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ELEN [110]
3 years ago
11

An archaelogist in Turkey discovers a

Mathematics
1 answer:
Travka [436]3 years ago
3 0

Step-by-step explanation:

hope this helps you to understand

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How to expand and simplify 5(3p+2)-2(5p-3)
Luda [366]
5(3p+2) - 2(5p-3)     Distribute the 5 and the 2
(5*3p) + (5*10) + (-2*5p) + (-2*-3)
15p + 50 - 10p + 6      Combine like terms
5p + 56
5 0
3 years ago
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If anyone knows about definite integrals for calculus then please I request help! I
kicyunya [14]

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:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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
  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

4 0
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The upper quartile of a set of data is the mode of the upper half of the data. <br> True<br> False
Arturiano [62]
I think it might be false, I could be wrong though. 
Hopefully that helped! :) 
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3 years ago
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CB = 4, CA = 11 , and CE = 8 , what is the length of overline ET ?
erma4kov [3.2K]
  • AB=11-4=7

\\ \rm\hookrightarrow \dfrac{BC}{AB}=\dfrac{CE}{ET}

  • ET be x

\\ \rm\hookrightarrow \dfrac{4}{7}=\dfrac{8}{x}

\\ \rm\hookrightarrow 4x=56

\\ \rm\hookrightarrow ET=x=14

8 0
2 years ago
How do i solve this (15 points)
GaryK [48]
Multiply 64 and 12 three times and 64 ×12=? then ?×3=?
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