I got Y=12 for my answer.
The standard form of the equation of a circle of radius r, with (assuming centre h, k) is given as:
(X-h)^2 + (y-k)^2 = r^2
As we are required to write an equation in standard form for the circle with radius 9 centred at the origin.
Centre(h,k)=(0,0), r=9
Substituting these values into the standard form of the equation of a circle given above:
(X-0)^2 + (y-0)^2 = 9^2
X^2 + y^2 =81
The standard form is x^2 + y^2 =81
I’m pretty sure this is right
Answer:
the ones with 16 dollar jeans
Answer
What is the question
Cause otherwise we can’t help


- <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

- <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>


According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.
<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>





<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>
The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse
