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Scilla [17]
3 years ago
8

The scheme where you can find the greatest common divisor of two integers by repetitive application of the division algorithm is

known as the Brady algorithm. True False
Mathematics
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

False

Step-by-step explanation:

Euclidean Algorithm is the algorithm that allows us to find the greatest common divisor (gcd) of two integers by repetitive application of the division algorithm.

A division algorithm is an algorithm which, given two integers N and D, computes their quotient and/or remainder.

Quotient and/or Remainder = \frac{N}{D}

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Geometry math question no Guessing and Please show work
MAXImum [283]

If the point P(x, y) partitions the segment AB in the ratio 1 : 1, then the point P is midpoint of segment AB.

The formula of a midpoint of segment:

A(x_A;\ y_A),\ B(x_B,\ y_B)\\\\P_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)

We have:

A(13,\ 1)\to x_A=13,\ y_A=1\\B(-5,\ -3)\to x_B=-5,\ y_B-3\\\\P\left(\dfrac{13+(-5)}{2};\ \dfrac{1+(-3)}{2}\right)\to P\left(\dfrac{8}{2};\ \dfrac{-2}{2}\right)\to P(4,\ -1)

Answer: C. (4, -1)

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4 years ago
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mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

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Step-by-step explanation:

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