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3 years ago

Find the missing angles for each problem, given that lines a and b are parallel (step by step)

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

Answer:

1) 40°

2) 140°

3) 140°

5)40°

6)140°

7)140°

8)40°

Step-by-step explanation:

1 = 40 ° [Vertically Opposite angles are equal]

8 = 40 ° [Corresponding Angles are equal]

5 = 8 = 40° [Corresponding Angles are equal]

7 = 180° - 40° = 140° [Supplementary angles sum 180°]

6 = 140° [Vertically Opposite angles are equal]

2 = 140° [Corresponding Angles are equal]

3 = 140 ° [Vertically Opposite angles are equal]

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Question is below thanks

vichka [17]

Answer:

33.51

Step-by-step explanation:

There's a formula using the value of the angle but I don't remember it so the most easy way is:

A circle has an angle of 360°

so our section has an angle of 60

and we know that 360=60*6

so the area of the circle we divide it by 6 and we get the area we want

$A=\pi r^2\\\\A=\pi (8)^2\\\\A=64\pi$

divide this by 6

$A_1=A/6=64\pi/6\approx 33.51$ units squared

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3 years ago

Read 2 more answers

Graph The Line. -2x+y= -6

loris [4]

0,6 !!

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3 years ago

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Hi, how do we do this question?

Nutka1998 [239]

Answer:

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Factoring

Algebra II

Polynomial Long Division

Calculus

Differentiation

Derivatives
Derivative Notation

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

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

Step 1: Define

Identify

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 3x + 1$
[u] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

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

Unit: Integration

Book: College Calculus 10e

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3 years ago

A human hair has a width of about 6.5 x 10-5 meter. What is this width written in standard notation?

kobusy [5.1K]

Answer:

The answer is C. 0.000065

Step-by-step explanation:

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4 years ago

On a diagram of the arena, the car park's length is 15 cm. The scale is 1:100.

faltersainse [42]

If a diagram has a scale of 1:100, that means that the real thing is 100 times larger. To find how long the real thing is, that means that we need to multiply the length on the diagram by 100. 100 * 15 is 1500. But, that is 1,500 cm. To convert it into meters, we need to divide by 100. That's because there's 100 centimeters in every meter. 1500/100 is 15, which means that the actual length of the car park is 15 meters.

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4 years ago