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:
0.609
Step-by-step explanation:
9.06
6.90
0.906
0.609
Answer: 0.609
The answer is 390km. You just multiply 130 x 3.
Answer:
P-E-M-D-A-S
Step-by-step explanation:
Ok, so start with what i call "P-E-M-D-A-S"
P: parentheses | Also known as these: ( )
E: exponents | Its a small number by another number, like the 3 on your screen
M: Multiply | Also Known as x or a dot
D: Divide | i forgot how to make it
A: Add | also known as +
S: Subtract | also known as -
now go in order to solve it, it makes it much more easy!
