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2 years ago

Algebra 1

Mathematics

1 answer:

Ymorist [56]2 years ago

4 0

Answer: $32

Step-by-step explanation:

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If angle 2 measures 155 degrees in the image below and what is the measurement of angle 1

Degger [83]

<h2>Answer:</h2>

<em>Angle 1 measures 25 degrees</em>

<h2>Explanation:</h2>

Hello! Remember you have to write complete questions in order to get good and exact answers. Here I'll assume the diagram shown below. In that figure, we know that:

$\angle 1=? \\ \\ \angle 2=155^{\circ}$

The figure shows that these two angles are supplementary, so they add up to 180 degrees. Therefore:

$\angle 1+\angle 2=180^{\circ} \\ \\ Substituting \ \angle 2=155^{\circ}: \\ \\ \angle 1+155^{\circ}=180^{\circ} \\ \\ \\ Isolating \ \angle 1: \\ \\ \angle 1=180^{\circ}-155^{\circ} \\ \\ \boxed{\angle 1=25^{\circ}}$

4 0

3 years ago

Write an explicit formula for an the nth term of the sequence 22,19,16

Georgia [21]

Answer:

Uₙ = 22 - 3(n - 1)

3 0

2 years ago

3. If mZEFH = (5x + 1), mZHFG = 62°, and<br> mZEFG = (18x + 11), find each measure.

arsen [322]

Answer/Step-by-step explanation:

Given:

m<EFH = (5x + 1)°

m<HFG = 62°

m<EFG = (18x + 11)°

Required:

1. Value of x

2. m<EFH

3. m<EFG

SOLUTION:

1. Value of x

m<EFH + m<HFG = m<EFG (angle addition postulate)

(5x + 1) + 62 = (18x + 11)

Solve for x using this equation

5x + 1 + 62 = 18x + 11

5x + 63 = 18x + 11

Subtract 18x from both sides

5x + 63 - 18x = 18x + 11 - 18x

-13x + 63 = 11

Subtract 63 from both sides

-13x + 63 - 63 = 11 - 63

-13x = -52

Divide both sides by -13

-13x/-13 = -52/-13

x = 4

2. m<EFH = 5x + 1

Plug in the value of x

m<EFH = 5(4) + 1 = 20 + 1 = 21°

3. m<EFG = 18x + 11

m<EFG = 18(4) + 11 = 72 + 11 = 83°

8 0

3 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$