Answer:
position: (-6, -4)
range: 6
Step-by-step explanation:
The equation is that of a circle centered at (-6, -4) with a radius of √36 = 6. We presume that the "position" is that of the circle's center, and the "range" is the radius of the circle.
___
The standard form equation of a circle with center (h, k) and radius r is ...
(x -h)^2 +(y -k)^2 = r^2
Matching parts of the equation, we find ...
h = -6, k = -4, r = √36 = 6.
For this case we have the following system of equations:

We observe that we have a quadratic equation and therefore the function is a parabola.
We have a linear equation.
Therefore, the solution to the system of equations will be the points of intersection of both functions.
When graphing both functions we have that the solution is given by:

That is, the line cuts the quadratic function in the following ordered pair:
(x, y) = (1, 2)
Answer:
the solution (s) of the graphed system of equations are:
(x, y) = (1, 2)
See attached image.
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
Join the centre O to the chord (let it be MN) & let OH be the perpendicular to the chord
OH bisects MN into 2 equal parts (each one is x/2)
OMH is a right triangle with one side =8, the second leg =x/2 & the hypotenuse = 12 (Radius)
Apply Pythagoras:
12² = 8² +(x/2)² ==>144=64 + x²/4 ==> x²=4(144-64) =320
x²=320==> x=√320 =17.88 ≈17.9