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3 years ago

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Mathematics

2 answers:

irinina [24]3 years ago

8 0

Answer:

triangle

Step-by-step explanation:

The cross section is a triangle.

Lera25 [3.4K]3 years ago

4 0

Answer:

i think either a square or rectangle

Step-by-step explanation:

You might be interested in

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Given f(x)=e^-x^3 find the vertical and horizontal asymptotes

Aleonysh [2.5K]

Given:

$f\mleft(x\mright)=e^{-x^3}$

To find the vertical and horizontal asymptotes:

The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.

But, here there is no such point.

Thus, the function f(x) doesn't have a vertical asymptote.

The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.

$\begin{gathered} y=\lim _{x\rightarrow\infty}e^{-x^3} \\ =e^{-\infty} \\ y=0 \\ y=\lim _{x\rightarrow-\infty}e^{-x^3} \\ y=e^{\infty} \\ =\infty \end{gathered}$

Thus, y = 0 is the horizontal asymptote for the given function.

5 0

1 year ago

How many 1/12pieces are in 3/3?

wel

Answer:

plz ion know what is the awnser to mine

Step-by-step explanation:

5 0

3 years ago

Help plz who ever helps they are nice

slava [35]

Im mean I’d be down to help

4 0

3 years ago

Read 2 more answers

Match the angle with it's corresponding classification

SVETLANKA909090 [29]

Answer:

∠TPR = right angle

∠SPQ = acute angle

∠QPT = obtuse angle

Step-by-step explanation:

Right angles are 90°, acute angles are angles less than 90°, obtuse angles are angles more than 90°

8 0

3 years ago