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4 years ago

Solve for x. |x| – 5 = 7

Mathematics

1 answer:

musickatia [10]4 years ago

6 0

Answer:

x=12 or x= -12 (answer A)

Step-by-step explanation:

|x| – 5 = 7

|x| = 12

x=12 or x= -12 (answer A)

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murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

6 0

3 years ago

Help me what’s the answer .

Mandarinka [93]

Answer:

x=3

Step-by-step explanation:

i helped my friend on this before but it was a word problem

8 0

3 years ago

The length of a rectangle is 6 meters more than its width. If the area of the rectangle is 135 square meters, find its dimension

solniwko [45]

We have the formula for the area of a triangle equal to A=l*w, where stands for the area, l for the length and w for the width. Given in the problem are the following:
A=135 m
l=6+w
w=w
Solving for the value of w, we have:
A=(w+6)*w
135=w²+6w
w=9 meters
Therefore, the length of the rectangle is 15 meters.

8 0

3 years ago

Plssss help!!! Asap tyyy.

Nana76 [90]

Answer:

1):Reaction

2):Action

3):Thought

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Step-by-step explanation:

5 0

2 years ago

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Your answer<br> What is the value of the expression? -50 + 51? *

bija089 [108]

Answer:

The answer is 1

Step-by-step explanation:

-50+50=1

3 0

3 years ago

Read 2 more answers