Qual o resultado de (-3+4-2)elevado a 5

Avaluate 2^3x-1 for x=1

DENIUS [597]

Answer:

4 or 7 (Check the explanation to see which one your equation actually is.

Step-by-step explanation:

One thing I want to clarify for you, It's Evaluate, not Avaluate.

Okay so, we want to find the value of 2^3x-1 and we know x = 1.

You didn't really clarify if the expression was $2^{3x-1}$ or $2^{3x}-1$ , so I'll be doing both:

For $2^{3x-1}$ , we should plug in x to get $2^{3(1)-1}$ and then simplify to get $2^{2}$ .

2 to the power of 2 or $2^{2}$ is equal to 2 * 2 or 4.

The second one, $2^{3x}-1$ , we should plug in x to get $2^{3(1)}-1$ and then to become $2^{3}-1$ . 2 to the power of 3 or $2^{3}$ is 8 and then minus 1 is 7.

7 0

3 years ago

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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$