PLEASE HELP IF YOU CAN!!!

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

8 0

Lol i forgot how to do this. Sorrryyy.

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Rewrite in Polar Form.... x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

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Why might the mean of a probability distribution for a discrete random variable be less than (or greater than) the average of po

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The Normal probability distribution function is left-skewed, right-skewed, or symmetric depending on the values of the variance and the standard deviation might the mean of a probability distribution for a discrete random variable be less than (or greater than) the average of possible values.

A probability distribution is a mathematical function that describes the probabilities of different possible values of a variable. Probability distributions are often represented using graphs or probability tables.

Probability distributions are called discrete probability distributions, and the set of outcomes is inherently discrete. For example, if you roll a die, all possible outcomes are discrete and you get a large number of outcomes. Also called probability mass function.

Learn more about probability distribution at

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