Pls help me whoo ever answers ill make u brainiest

Mathematics

2 answers:

pogonyaev3 years ago

7 0

Bro just type the equations in on the calculator, it should give you the answers.. this too easy bro. Stop being lazy

11Alexandr11 [23.1K]3 years ago

7 0

Please me anyway you can just look it up on a calculator

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If I have 150,000 crystals, and ONE costume cost 4,000 crystals, how many costumes can I buy total?

MArishka [77]

Okay, think of it this way:

You have to find how many times 4,000 goes into 150,000.

This will tell you how many costumes you can buy.

To make this easier, we can divide 4 by 150. (this is proportional to 4,000 and 150,000 since we took 3 zeros from each number; you will get the same answer)

150/4= 37.5

You can buy 37 whole costumes.

Hope this helped!

6 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is