All categories

3 years ago

Is this right? I’m confused for answering either B or C.

Mathematics

2 answers:

Stells [14]3 years ago

4 0

I think that the answer is C.

hope it helps :)

boyakko [2]3 years ago

3 0

your profile says you're 22 but I find that painfully challenging to believe. She knits 5 scarves in 2 hours. Divide 10 by 2, you get 5. 5x5 is 25. It's C. You shouldn't need help. This is sad.

You might be interested in

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

umka2103 [35]

Replace $x\mapsto \tan^{-1}(x)$ :

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx$

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace $x\mapsto\frac1x$ :

$\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx$

Then the original integral is equivalent to

$\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx$

Recall that for |x| < 1,

$\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}$

so that we can expand the integrand, then interchange the sum and integral to get

$\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx$

Integrate by parts, with

$u = \ln(x) \implies du = \dfrac{dx}x$

$du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

Recall the Fourier series we used in an earlier question [27217075]; if $f(x)=\left(x-\frac12\right)^2$ where 0 ≤ x ≤ 1 is a periodic function, then

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)$

Evaluate f and its Fourier expansion at x = 1/2 :

$\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)$

$\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}$

So, we conclude that

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}$

3 0

2 years ago

Solve for x.. please help

Genrish500 [490]

The answer for this problem is that x = 60°, or B.

We can solve this problem by using our knowledge of quadrilaterals. In this problem, we need to know that the number of degrees in a quadrilateral is equal to 360, but, the formula to find the number of degrees in a polygon with n sides is (n-2) * 180.

Now that we know that the sum of the angles if 360°, we can add all the angles and solve for x. The angles with the squares are 90°.

2x° + x° + 90° + 90° = 360°

Get x by itself, combine like terms.

3x° + 180° - 180° = 360° - 180°
3x° = 180°
x = 60°, or B.

4 0

4 years ago

Read 2 more answers

Describe the x-values for which (a) f is increasing or decreasing, (b) f(x) > 0 and (c) f(x) < 0

Kryger [21]

(a) The function "f" is increasing for the x-values of [(-∞) < x ≤ 0] and [2 < x < ∞], while "f" is decreasing for the x-values of [0 < x ≤ 2].

(b) The function f(x) is greater than 0 at [(-1) < x < ∞], excluding the point (x = 2)

In the reference attachment attached along with the question statement, we are provided with a graph of a certain function "f",

And as per the question statement, we are required to determine the x-values for the following:

(a) for which "f" is increasing or decreasing,

(b) [f(x) > 0]

From the graph, it is very clear that, the slope is rising from [(-∞) < x ≤ 0] and [2 < x < ∞], while, declining from [0 < x ≤ 2] of the x-axis. Therefore,

(a) The function "f" is increasing for the x-values of [(-∞) < x ≤ 0] and [2 < x < ∞], while "f" is decreasing for the x-values of [0 < x ≤ 2].

(b) The function f(x) is greater than 0 at [(-1) < x < ∞], excluding the point (x = 2)

To learn more about x-values of increasing or decreasing functions, click on the link below.

brainly.com/question/28912753

#SPJ1

7 0

1 year ago

This is my last question I need to do what’s the answer

MAXImum [283]

Step-by-step explanation:

According to the question,

l+5=2h

5 0

3 years ago

Hii please help i’ll give brainliest!!

Valentin [98]

Answer:

friction

lol you’re welcome

3 0

3 years ago