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3 years ago

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Mathematics

1 answer:

kramer3 years ago

7 0

Answer:

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Step-by-step explanation:

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What polygon is shown below?

HACTEHA [7]

That shape is a Rhombus

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3 years ago

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I have a bag of trail mix. One half of the bag is peanuts 1/4 of the bag is chocolate candies,1/4 of the bag is dried fruit. Wha

lana66690 [7]

50% or 1/2 or 1:2 would be your answer

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3 years ago

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

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2 years ago

How long does it take to drive 341 miles at 63 miles per hour?

Diano4ka-milaya [45]

If you maintained an average speed of 63 mph, the time to drive 341 miles would be about 5 hours and 24 minutes. You would have to add time for stops, detours and other delays.

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3 years ago

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This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

harina [27]

$L(x,y,z,\lambda)=8x+8y+4z+\lambda(4x^2+4y^2+4z^2-36)$

$L_x=8+8\lambda x=0\implies 1+\lambda x=0$
$L_y=8+8\lambda y=0\implies 1+\lambda y=0$
$L_z=4+8\lambda z=0\implies 1+2\lambda z=0$
$L_\lambda=4x^2+4y^2+4z^2-36=0\implies x^2+y^2+z^2=9$

$yL_x=y+\lambda xy=0$
$xL_y=x+\lambda xy=0$
$\implies yL_x-xL_y=y-x=0\implies y=x$

$2zL_x=2z+2\lambda xz=0$
$xL_z=x+2\lambda xz=0$
$\implies 2zL_x-xL_z=2z-x=0\implies x=2z$

$2zL_y=2z+2\lambda yz=0$
$yL_z=y+2\lambda yz=0$
$\implies 2zL_y-yL_z=2z-y=0\implies y=2z$

$x=y=2z\implies x^2+y^2+z^2=9\iff 4z^2+4z^2+z^2=9z^2=9\implies z=\pm1$

$z=\pm1\implies y=x=\pm2$

So we have two critical points, (2, 2, 1) and (-2, -2, -1), which respectively give a max of 36 and a min of -36.

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3 years ago