Let's try to complete the squares.
The x-part starts with
, which is the beginning of
. So, we'll think of
as ![(x+5)^2-25](https://tex.z-dn.net/?f=%28x%2B5%29%5E2-25)
Similarly, we have that
![y^2+12y = (y+6)^2-36](https://tex.z-dn.net/?f=y%5E2%2B12y%20%3D%20%28y%2B6%29%5E2-36)
So, the equation becomes
![x^2 + y^2 + 10x + 12y + 25 = 0 \iff (x+5)^2-25 + (y+6)^2-36+25=0 \iff (x+5)^2+ (y+6)^2-36=0 \iff (x+5)^2+ (y+6)^2=36](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20%2B%2010x%20%2B%2012y%20%2B%2025%20%3D%200%20%5Ciff%20%28x%2B5%29%5E2-25%20%2B%20%28y%2B6%29%5E2-36%2B25%3D0%20%5Ciff%20%28x%2B5%29%5E2%2B%20%28y%2B6%29%5E2-36%3D0%20%5Ciff%20%28x%2B5%29%5E2%2B%20%28y%2B6%29%5E2%3D36)
Now we have writte the equation of the circle in the form
![(x-k)^2+(y-h)^2=r^2](https://tex.z-dn.net/?f=%28x-k%29%5E2%2B%28y-h%29%5E2%3Dr%5E2)
When the equation is in this form, everything is more simple: the center is
and the radius is
.
Answer:
44.7 mg
Step-by-step explanation:
The equation for exponential decay can be written in the form ...
y = a·b^(t/p)
where 'a' is the initial value, 'b' is the decay factor, 'p' is the period over which the decay factor is applicable, and t is time in the same units as p.
<h3>Setup</h3>
Using the above equation, we have ...
a = initial value = 110 mg
b = decay factor = 55/110 = 1/2 over time period p=20 hours
Then the equation is ...
y = 110·(1/2)^(t/20) . . . . amount remaining after t hours
<h3>Solution</h3>
We want the amount remaining after 26 hours. That will be ...
y = 110·(1/2)^(26/20) ≈ 44.67
About 44.7 milligrams will remain after 26 hours.
Answer: 15
Explanation: quick maths in the brain
Answer:
y = mx + 1/2
Step-by-step explanation:
(x,y)
- Plot each point on a graph
- The line should go from the bottom left (-6,-10) to the top right (1,4)
- Count how many spaces are on the y-axis from -6 to 1 for numerator
- Count how many spaces are on the x-axis from -10 to 4 for denominator
- Positive slope
- Slope = 7/14
- Simplify to 1/2