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LiRa [457]
3 years ago
11

Which mathematical property is demonstrated below? 3 + 0 = 3

Mathematics
2 answers:
Lera25 [3.4K]3 years ago
6 0

Answer: Identity Property

Step-by-step explanation:

The rest have very fake names lol

Wittaler [7]3 years ago
6 0

Answer:

Identity Property

Step-by-step explanation:

Identity Property :

Any number plus zero is the original number.

<h3>Hope it is helpful....</h3>
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Where is the center of the largest circle that you could draw inside a given triangle? . A.the point of concurrency of the media
juin [17]

The <em><u>correct answer</u></em> is:

C.the point of concurrency of the angle bisectors of the triangle

Explanation:

The largest circle that can be drawn inside a triangle is called an inscribed circle.  The center of this circle is called the incenter.

The incenter is formed by the intersection of the angle bisectors of all 3 angles in the triangle.

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-9 is____ units away from 0 on the number line.
Aloiza [94]

Answer:

My answer would be 9 unit away

Step-by-step explanation:

becuase you can subtract 0 from -9

0 + -9  would equal -9

then 9 - (-9) = 0

so 9 is my answer

sorry if i am wrong

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Witch of the following numbers is clostest to 6?
Reika [66]
3,2,1,6 are all factors of 6.
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Given f(x), match each expression to its correct value
Murljashka [212]

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-11, f(-5)    1

Step-by-step explanation:

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2 years ago
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The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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