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dsp73
1 year ago
10

I will give brainliest to the fastest helpful answer! :) A figure is located at (0, 0), (−3, −4), and (−3, 0) on a coordinate pl

ane. What kind of 3-D shape would be created if the figure was rotated around the x-axis? Provide an explanation and proof of your answer to receive full credit. Include the dimensions of the 3-D shape in your explanation.
Mathematics
1 answer:
yulyashka [42]1 year ago
5 0

The 3-D shape would be created if the figure was rotated around the x-axis is a cone

<h3>What are 3-D shapes?</h3>

3-D shapes (short form of 3-Dimensional shapes) are shapes that have width, length and height

<h3>How to determine the 3-D shape?</h3>

The coordinates are given as:

(0, 0), (-3, -4) and (-3, 0)

When the above coordinates are plotted on a coordinate plane and the points are connected;

We can see that the points form a right-triangle

See attachment for the shape

As a general rule

Rotating a right-triangle across the x-axis would form a cone

Hence, the 3-D shape would be created if the figure was rotated around the x-axis is a cone

Read more about rotation at:

brainly.com/question/4289712

#SPJ1

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A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
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so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
18 &gt; s-(-4) please help
coldgirl [10]
I hope this helps you

7 0
3 years ago
Read 2 more answers
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