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.
Answer:
Step-by-step explanation:
is the same as 4/3 so you'd be multiplying 4/3 x 4/3 so you multiply straight across to get 16/9 when you simplify you are left with 1 7/9
Hope that helps and have a great day!
Answer:
can I get a hiaaa
How was the government able to make children go to residential school?
Divide 294 by 14 and you get 21. I hope it helps and that’s the answer. :)
Y=mx+b
slope=m
yint=b
we are given
slope=-2/3
point=(-3,-1)
y=-2/3x+b
input point -3,-
x=-3
y=-1
find b
-1=-2/3(-3)+b
-1=2+b
minus 3
-3=b
y=-2/3x-3
answer is D
