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3 years ago

I need help with this

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

Answer:

512

Step-by-step explanation:

64 × 8 = 512

512

64 8

16 4 2

8 2 2 1

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1) What can u do to get 2x alone on the left side?

CaHeK987 [17]

Answer:

82.5 - 82.5 + 2x = 338.5 - 82.5

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2x ÷ 2 = 256 ÷ 2

x = 128

Step-by-step explanation:

In order to get rid 82.5, you have to substract the same value on both side to get a 0 value.

In order to find the value of x, you have to divide the value that is sticked with x which is 2. As you divide 2 by 2, you will get 1 as a single value of x so you have to divide 2 to both side too.

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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