Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
You have to complete the square on this to get it into standard form of a circle. Move the 8 over to the other side because that's part of the radius. Group together the x terms, take half the linear term which is 8, square it and add it in to both sides. Half of 8 is 4, 4 squared is 16, so add in 16 to both sides. I'll show you in a sec. You don't need to do anything to the y squared term. This just means that the center of the circle does not move up or down, only side to side, right or left. Here's your completing the square before we simplify it down to its perfect square binomial.

. Now break down the parenthesis into the perfect square binomial and do the addition of the right:

. This is the standard form of a circle that has a center of (4, 0) and a radius of
Answer:

Step-by-step explanation:

Formula :


<h3>Hope it is helpful...</h3>
Minor angle, The arc it produces is less than 180 degrees so it is a minor angle
Find the mean, median, and mode of 14, 15, 3, 15, 14, 14, 18, 15, 8, 16.
klemol [59]
Answer:
mean: 13.2 (average)
median: 14.5 (Center)
mode: 14 and 15 (because data is bimodal so there are two modes
Step-by-step explanation: