Is 13 a perfect square

1 answer:

8 0

$1^2=1 \\ 2^2=4 \\ 3^2=9 \\ 4^2=16 \\ 5^2=25 \\ 6^2=36 \\ 7^2=49 \\ 8^2=64 \\ 9^2=81 \\ 10^2=100 \\ 11^2=121 \\ 12^2=144 \\ \boxed{13^2=169} \\ 14^2=196 \\ 15^2=225 \\$

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denis-greek [22]

X is greater than 4

3 0

3 years ago

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

skad [1K]

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

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

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

3 0

3 years ago

Huddson has practiced a 360-back flip for months. This week, he did it perfectly 24 out of 54 times. At this rate, how many time

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