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3 years ago

Turn -6 13/15 as a decimal

Mathematics

1 answer:

barxatty [35]3 years ago

7 0

-6.867
You do 13 divided by 15 the plug in the -6 in the front

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Identify the domain of the function.<br> (-4,4)<br> [-4,4)<br> (-4,4)<br> (-4, 4)

Vlad1618 [11]

All real numbers I think

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3 years ago

Mrs. Crimm signed up to bring two dozen cupcakes for the school bake sale. She made one dozen, but her children ate 3 of the cup

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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:

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

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

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

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

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

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

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

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

4 0

1 year ago

Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at

Dafna1 [17]

Answer:

$\huge\boxed{ \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1}}}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

conic sections
PEMDAS

<h3>tips and formulas:</h3>

$\sf hyperbola \:equation : \\ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1$
vertices of hyperbola:(±a,0) and (0,±b) if reversed
$\sf \: asymptotes : \\ y = \pm\frac{b}{a} x$