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stepan [7]
3 years ago
14

Which is a true statement about an exterior angle of a triangle?

Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
5 0

Solution:

we have been asked to find the true statement about the exterior angle from the given statements.

As we know that the exterior angles are always out of the triangle and it makes a sum of 180 with anyone of the internal angle of the triangle.

The pair lies on the same straight line and makes a pair.

Hence we can say that it is formed by a linear pair with one of the interior angles of the triangle.

Hence the correct option is D.

8090 [49]3 years ago
4 0
D. It forms a linear pair with one of the interior angles if the triangle. This is 100% right im taking the test rn and i got it right.yw
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ladessa [460]

Answer: <u>11.</u>

Step-by-step explanation: All you'd have to do in order to achieve this answer is by dividing 410 by 40 which gives you the answer of 10.25. In order to fit everyone they'll need a rounded number of buses; equalling 11.

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Step-by-step explanation:

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Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

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