3 years ago

How do I do the additional names part

Mathematics

1 answer:

zubka84 [21]3 years ago

6 0

Equilateral quadrilateral

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Elvaluate 3^-2 jgigjdkcjcivkxjgfnv

BARSIC [14]

Answer:

1/9

Step-by-step explanation:

when you have a negative exponent, you move the term to the denominator: 1/3^2

then you just do 3^2: 1/9

4 0

4 years ago

Sin(x2 y2 da r , where r is the region in the first quadrant between the circles with center the origin and radii 1 and 5

Rudik [331]

I assume there's a plus sign missing above...

Convert to polar coordinates, using

$\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}$

Then the Jacobian is

$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}x_r&y_r\\x_\theta&y_\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&\sin\theta\\r\sin\theta&-r\cos\theta\end{vmatrix}=-r$

Then

$\mathrm dA=\mathrm dx\,\mathrm dy=|-r|\,\mathrm dr\,\mathrm d\theta=r\,\mathrm dr\,\mathrm d\theta$

so the integral can be written as

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_1^5r\sin(r^2)\,\mathrm dr\,\mathrm d\theta$

Let $s=r^2$ , so that $\dfrac{\mathrm ds}2=r\,\mathrm dr$ .

$\displaystyle\frac12\int_0^{\pi/2}\int_1^{25}\sin s\,\mathrm ds\,\mathrm d\theta=-\frac12(\cos25-\cos1)\int_0^{\pi/2}\mathrm d\theta=\frac\pi4(\cos1-\cos25)$

3 0

3 years ago

Verify the identity sin^2(x/2)=(1-cosx)/2

horsena [70]

Use the following identity:

cos(2x)=1-2sin^2(x)

5 0

4 years ago

Simplify the following without a calculator leave your answer with positive exponents

alexira [117]

It would be simplified: 2/375

8 0

3 years ago