3(y+2)+2(4-3)=-15 please help

Mathematics

1 answer:

trasher [3.6K]2 years ago

7 0

3y+6+8-6=-15
3y+8=-15
3y=-15-8
3y=-23
y=23/3 or 7.6

HOPE THIS HELP!!! :D

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The hypotenuse of a right triangle is four times the length of one of the legs. The length of the other leg is 1500⎯⎯⎯⎯⎯⎯⎯⎯√ fee

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Step-by-step explanation:

So this is just playing with the pythagorean theorem

a^2 + b^2 = c^2 with a and b as legs and c as the hypotenuse.

Let's say a is equal to 1500 so then c is 4 times b

1500^2 + b^2 = (4b)^2 Now we just treat this like a normal algebraic expression

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sqrt(150,000) = b

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You can check this now.

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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