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3 years ago

HEY GUYS HELP, ILL MARK YOU BRAINLY !

Mathematics

1 answer:

Serjik [45]3 years ago

5 0

Y=40x+100 and the second question is y=1.69x

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Find the slope and the y-intercept of 3x-6y=48

Sedaia [141]

Answer:

Step-by-step explanation:

3x-6y=48

Turn the equation into slope intercept form (y=mx+b)

x-2y=16 (Subtract x)

-2y=16-x (Divide by -1/2)

y=1/2x-8

The y-intercept is b, so -8 is the y-intercept.

The slope is m, so the slope is 1/2.

3 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

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

Subtraction and division are derived from which principal operations

JulsSmile [24]

There is no such thing as subtraction. Just something to think about :)

There are numbers and there is an operation called addition. Every number has an additive inverse, an opposite value.

Example: The opposite of 4 is -4.

When we do addition between numbers like 5 and the inverse of 2: 5 + -2 this is usually what we're thinking of when we talk about subtraction. We could write it as 5 - 2.

But anyway, to answer your question,

Subtraction and division are derived from the principle operations of ADDITION and MULTIPLICATION.

7 0

4 years ago

If there are 25 out of 65 people who are wearing red at a christmas party, what percent is of the people are wearing red

yuradex [85]

25 out of 65 people are wearing red, if there are 65 people at the party and only 25 are wearing red you would find the percentage by dividing 65 by 25 which equals to 2.6%

i hope this is correct im not great at math but i hope this helps

3 0

3 years ago

What 175 multiplyed by 260

jolli1 [7]

Answer:

45500

Step-by-step explanation:

175 can be multiplied by 200 which gives you 35000 plus 175 times 60 which gives you 10500. You can add and you get 45500.

I was a first place champion in number sense and thats my strategy

Thank you!!!

May i have brainliest?

6 0

3 years ago