What is the slope? (1,7), (10,1)

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ $m= \frac{1-7}{10-1}$ $m= -\frac{2}{3}$

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Convert: 6y + y² = x² from rectangular to polar form.

11Alexandr11 [23.1K]

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\\\
cos(2\theta)=
\begin{cases}
\boxed{cos^2(\theta)-sin^2(\theta)}\\
1-2sin^2(\theta)\\
2cos^2(\theta)-1
\end{cases}\\\\\\
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-------------------------------\\\\$

$\bf 6y+y^2=x^2\implies \cfrac{6y}{x^2}+\cfrac{y^2}{x^2}=1\implies \cfrac{6rsin(\theta )}{[rcos(\theta )]^2}+\cfrac{[rsin(\theta )]^2}{[rcos(\theta )]^2}=1
\\\\\\
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$\bf \cfrac{6sin(\theta )}{rcos^2(\theta )}=1-\cfrac{sin^2(\theta )}{cos^2(\theta )}\implies \cfrac{6sin(\theta )}{1-\frac{sin^2(\theta )}{cos^2(\theta )}}=rcos^2(\theta )
\\\\\\
\cfrac{6sin(\theta )}{cos^2(\theta )\left[ 1-\frac{sin^2(\theta )}{cos^2(\theta )} \right]}=r\implies \cfrac{6sin(\theta )}{cos^2(\theta )-sin^2(\theta )}=r
\\\\\\
\cfrac{6sin(\theta )}{cos(2\theta )}=r$

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3 years ago

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8.9 x 10^-10 in standard form

Harlamova29_29 [7]

Answer:

science E-notation: 8.900e-10