Which of the following is equalvalit to the expression shown, 4x + 3(2x - 1) - 4x

2 answers:

6 0

A.) 6x - 3

Explanation: the two 4’s on both sides of the equation cancel out which leaves you with 3 (2x-1). You use distributive property to distribute the 3 into the brackets which leads you with 6x-3. Hope it helped!

Sladkaya [172]4 years ago

5 0

The answer is A. 6x - 3

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Use the given transformation to evaluate the integral. double integral 9xy dA R , where R is the region in the first quadrant bo

lianna [129]

It looks like the boundaries of $R$ are the lines $y=\dfrac23x$ and $y=3x$ , as well as the hyperbolas $xy=\frac23$ and $xy=3$ . Naturally, the domain of integration is the set

$R = \left\{(x,y) ~:~ \dfrac{2x}3 \le y \le 3x \text{ and } \dfrac23 \le xy \le 3 \right\}$

By substituting $x=\frac uv$ and $y=v$ , so $xy=u$ , we have

$\dfrac23 \le xy \le 3 \implies \dfrac23 \le u \le 3$

and

$\dfrac{2x}3 \le y \le 3x \implies \dfrac{2u}{3v} \le v \le \dfrac{3u}v \implies \dfrac{2u}3 \le v^2 \le 3u \implies \sqrt{\dfrac{2u}3} \le v \le \sqrt{3u}$

so that

$R = \left\{(u,v) ~:~ \dfrac23 \le u \le 3 \text{ and } \sqrt{\dfrac{2u}3 \le v \le \sqrt{3u}\right\}$

Compute the Jacobian for this transformation and its determinant.

$J = \begin{bmatrix}x_u & x_v \\ y_u & y_v\end{bmatrix} = \begin{bmatrix}\dfrac1v & -\dfrac u{v^2} \\\\ 0 & 1 \end{bmatrix} \implies \det(J) = \dfrac1v$

Then the area element under this change of variables is

$dA = dx\,dy = \dfrac{du\,dv}v$

and the integral transforms to

$\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \int_{\sqrt{2u/3}}^{\sqrt{3u}} \frac{dv\,du}v$

Now compute it.

$\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \ln|v|\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u}} \,du \\\\ ~~~~~~~~ = \int_{2/3}^3 \ln\left(\sqrt{3u}\right) - \ln\left(\sqrt{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln(3u) - \ln\left(\frac{2u}3\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln\left(\frac{3u}{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \int_{2/3}^3 du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \left(3-\frac23\right) = \boxed{\frac76 \ln\left(\frac92\right)}$