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4 years ago

Write words to match the expression 3+(4×12)

2 answers:

5 0

The process of this problem is that you always do the equation that's in parentheses first so four multiplied by twelve equals fourtyeight. Fourtyeight plus three equals one hundred and fourty four.

Airida [17]4 years ago

3 0

3 more than the produce of 12 and 4.

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3 years ago

Which of the following is Th e equation of the boundary line shown in the following inequality?

In-s [12.5K]

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The answer is C.

Step-by-step explanation:

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3 years ago

xeze [42]

Answer:

Step-by-step explanation:

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3 years ago

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Temka [501]

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Through a given point there passes: one and only one line perpendicular to a plane. one and only one plane perpendicular to a line.

Step-by-step explanation:

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4 years ago

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

4 years ago