The area of the base b is the area of a square with sides of length 230 m. so, v =

1 answer:

4 0

Not completely sure what you are asking, but the area of a square with the side lengths of 230m would be 230*230=52,900.

You might be interested in

Questions. multiple choicw

jeyben [28]

11. 8 * 18 = 144 in³, Option D

12. Half the height of 8 cm is 4 cm. Volume = 4 * 6 * 10 = 240 cm³, Option D

13. Double the dimensions you get 8 cm for the height, 6 cm for the radius. Then plug in. V = pi * (6)² * 8 >> pi * 36 * 8 = 288pi or ≈ 904.78, Option C

14. Half of all the dimensions are 1 in, 4 in, and 3 in. 1 * 4 * 3 = 12 in³, Option B

15. $\frac{10}{16} = \frac{x}{18}$ >> 16x = 180 >> x = 11.25 so Option B

16. Option D, 10 cm.

17. Option C, 8.5 in

18. Option B. 10.2 km

19. Option D. 0.82

20. cos 30 = b / 11.5 >> b = 11.5(cos (30)) = 9.96 m, Option C.

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