Find the slope of the line. Use the "/" to make a fraction if needed. Reduce your answer.

Mathematics

1 answer:

blsea [12.9K]2 years ago

6 0

Answer:

$-\frac{1}{2}$

Step-by-step explanation:

This line goes down 1 unit and right 2 units.

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Can someone help me please I need the mode, median, range and the mean.?

Neko [114]

Answer:

First the mode. Since 5 popped up the most, 5 is the mode.

Next is the median. I crossed 1 dot from each side until it shows the last dot, and 5 was the last one.

After that the range. 9-2=7

Finally the worst, the mean... 2+2+3+3+3+4+5+5+5+5+5+6+6+6+8+9+9+9+9

=104/19=5.47

SO, Mode=5 Median=5 Range=7, and the mean is 5.47 (rounded nearest hundred)

7 0

2 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