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alexgriva [62]
3 years ago
14

HELP PLEASE!!

Mathematics
1 answer:
seropon [69]3 years ago
3 0

Answer:

BD = <u>1</u> unit

AD = <u>1</u> unit

AB = <u>1.6</u> units

AC = <u>1.6</u> units

Step-by-step explanation:

In the picture attached, the triangle ABC is shown.

Given that triangle ABC is isosceles, then ∠B = ∠C

∠A + ∠B + ∠C = 180°

36° + 2∠B = 180°

∠B = (180° - 36°)/2

∠B = ∠C  = 72°

From Law of Sines:

sin(∠A)/BC = sin(∠B)/AC = sin(∠C)/AB

(Remember that BC is 1 unit long)

AB = AC = sin(72°)/sin(36°) = 1.6

In triangle ABD, ∠B = 72°/2 = 36°, then:

∠A + ∠B + ∠D = 180°

36° + 36° + ∠D = 180°

∠D =  180° - 36° - 36° = 108°

From Law of Sines:

sin(∠A)/BD = sin(∠B)/AD = sin(∠D)/AB

(now ∠A = ∠B)

BD = AD = sin(∠A)*AB/sin(∠D)

BD = AD = sin(36°)*1.6/sin(108°) = 1

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Answer:

term a is 2x²

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Step-by-step explanation:

2x² + 24 = -4x

2x² + 4x + 24 = 0

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term b is 4x

term c is 24

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Answer:

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Step-by-step explanation:

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Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.
Maru [420]

Answer:

9\text{ln}|x|+2\sqrt{x}+x+C

Step-by-step explanation:

We have been an integral \int \frac{9+\sqrt{x}+x}{x}dx. We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx

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Now, we will apply the sum rule of integrals as:

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Using common integral \int \frac{1}{x}dx=\text{ln}|x|, we will get:

9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx

Now, we will use power rule of integrals as:

9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx

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We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

9\text{ln}|x|+2\sqrt{x}+x+C

Therefore, our required integral would be 9\text{ln}|x|+2\sqrt{x}+x+C.

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