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2 years ago

If you move point B along BC how does M

Mathematics

2 answers:

Nezavi [6.7K]2 years ago

7 0

Answer:

Step-by-step explanation:

How does M what?

Vladimir79 [104]2 years ago

4 0

Answer:

how does M what

Step-by-step explanation:

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How do u write any number to the zeroth power

SSSSS [86.1K]

Answer:

0 as an exponent would look something like 6^0 and anything to the power of 0 is always 1

7 0

2 years ago

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

X2 2 ab ab to b The diagian K n shows rectangle. The shade Find x.

Norma-Jean [14]

Answer:

x = 28 cm

Step-by-step explanation:

Given:

Area of link shaded regions = 84 cm²

Required:

The value of x (diameter of the semicircle/length of the rectangle)

Solution:

Diameter of the semicircle = 2r = x

Length of rectangle (L) = 2r = x

Radius of semicircle (r) = ½x

Width of rectangle (W) = radius of semicircle = ½x

Use 3.14 as π

Area of the link shaded regions = area of rectangle - area of semicircle

Thus:

Area of the link shaded regions = (L*W) - (½*πr²)

Plug in the values

84 = (x*½x) - (½*3.14*(½x)²)

84 = x²/2 - (1.57*x²/4)

84 = x²(½ - 1.57/4)

84 = x²(0.5 - 0.3925)

84 = x²(0.1075)

Divide both sides by 0.1075

84/0.1075 = x²

781.4 = x²

√781.4 = x

27.9535329 = x

x = 28 cm

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3 years ago

The length of a rectangle is 3 more than five times its width. If the perimeter of

AleksandrR [38]

Answer:

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To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides. This number divided by 2 is the width.

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2 years ago

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3 years ago

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