Using the equation of the circle, it is found that since it reaches an identity, the point (√5, 12) is on the circle.
<h3>What is the equation of a circle?</h3>
The equation of a circle of center 
and radius r is given by:
In this problem, the circle is centered at the origin, hence 
.
The circle contains the point (-13,0), hence the radius is found as follows:
Hence the equation is:
Then, we test if point (√5, 12) is on the circle:
25 + 144 = 169
Which is an identity, hence point (√5, 12) is on the circle.
More can be learned about the equation of a circle at brainly.com/question/24307696
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Answer:
4,832 yards
Step-by-step explanation:
302x16=
Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:
A)R1-R3
![\left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%260%7C0%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
B)2R2-R3
![\left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C-2%262%261%7C4%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
C)-2R1+R3
![\left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-1%7C-1%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
D)2R1+R3
![\left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%260%263%7C3%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
E)3R1+R3
![\left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%260%264%7C4%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
None of the options will triangularize this matrix. The only way to <span>triangularize this matrix is
R3-2R1
</span>
![\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C0%261%261%7C6%5C%5C0%260%26-1%7C-1%5Cend%7Barray%7D%5Cright%5D%20%20)
<span>
This equation is similar to C) but in reverse order. Order in which rows are written is important.</span>
