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3 years ago

25. what is happening in the diagram (5 points)

Mathematics

2 answers:

nexus9112 [7]3 years ago

7 0

You need to give more info.

Furkat [3]3 years ago

6 0

I don’t understand give more information

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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:

Calculus

Differentiation

Derivatives
Derivative Notation

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

Basic Power Rule:

f(x) = cxⁿ
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:

Step 1: Define

Identify

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

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

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

Step 3: Integrate Pt. 2

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

2 years ago

I will give the brainliest answer

GuDViN [60]

Answer:

$82 {units}^{2}$

Step-by-step

explanation

explanation is attached to the picture

hope this helps

brainliest appreciated

good luck! have a nice day!

3 0

3 years ago

Read 2 more answers

What is the area of this figure

anastassius [24]

Answer:

185

Step-by-step explanation:

sorry. had to do something but im bsck :)

5 0

3 years ago

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Five less than the product of 3 and a number of 40 . Write in an equation

vagabundo [1.1K]

Answer:

I don't get the "number of 40" part, but here's my best shot!

3(40x)-5

Step-by-step explanation:

4 0

3 years ago

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A cylinder and a cone have the same base and height. The cylinder can hold about 4,712 mc015-1.jpg of sand. Jared says that the

exis [7]

The correct answer is C. So, lets go first to the equations that determine the volume of these solids. For the cylinder, that is simple; pi*r*r*h where r is the radius, h is the height and pi is the constant 3.14. For the cone, it is not easy to derive but one gets that the formula is: $\frac{ \pi*r^2*h}{3}$ .

We notice thus that $V_{cyl}=\pi*r^2*h=3V_{cone}$ . That holds irrespective of their radius or height; we only need to know that the heights and radii of the two objects are the same. Now, we have thus that:
$\frac{V_{Cyl}}{V_{Cone} } =1/3$ .

We can check if this holds for Jared's statement; 1178/4712=0.25=1/4. So, it does not hold and thus Jared's statement is incorrect.

6 0

3 years ago

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