A. but I would double check on that one first.
Answer:
Hey there!
3 x 5/4 = 3 x 5 ÷ 4 = 5 x 3 ÷ 4
5 x 4/3 = 5 x 4 ÷ 3 = 4 x 5 ÷ 3
4 x 3/5 = 4 x 3 ÷ 5 = 4 x 3 ÷ 5
Hope this helps :)
Answer:
p = 4
Step-by-step explanation:
Given equation:
![x^2+(p-3)y^2-4x+6y-16=0](https://tex.z-dn.net/?f=x%5E2%2B%28p-3%29y%5E2-4x%2B6y-16%3D0)
<u>Standard equation of a circle:</u>
![(x-a)^2+(y-b)^2=r^2](https://tex.z-dn.net/?f=%28x-a%29%5E2%2B%28y-b%29%5E2%3Dr%5E2)
(where
is the centre of the circle, and
is the radius)
If you expand this equation, you will see that the coefficient of
is always one.
Therefore, ![p-3=1](https://tex.z-dn.net/?f=p-3%3D1)
![\implies p=1+3=4](https://tex.z-dn.net/?f=%5Cimplies%20p%3D1%2B3%3D4)
<u>Additional information</u>
To rewrite the given equation in the standard form:
![\implies x^2+y^2-4x+6y-16=0](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2%2By%5E2-4x%2B6y-16%3D0)
![\implies x^2-4x+y^2+6y=16](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2-4x%2By%5E2%2B6y%3D16)
![\implies (x-2)^2-4+(y+3)^2-9=16](https://tex.z-dn.net/?f=%5Cimplies%20%28x-2%29%5E2-4%2B%28y%2B3%29%5E2-9%3D16)
![\implies (x-2)^2+(y+3)^2=16+4+9](https://tex.z-dn.net/?f=%5Cimplies%20%28x-2%29%5E2%2B%28y%2B3%29%5E2%3D16%2B4%2B9)
![\implies (x-2)^2+(y+3)^2=29](https://tex.z-dn.net/?f=%5Cimplies%20%28x-2%29%5E2%2B%28y%2B3%29%5E2%3D29)
So this is a circle with centre (2, -3) and radius √29
You should get photo math you can take a picture and it will give you the answer and show you how to solve it