The equation of a circle is written as (x-h)^2 +(y-k)^2 = r^2
h is the x value of the center and k is the y value of the center and r is the radius.
Replace h, k and r with the given information:
(x- -7)^2 + (y - -6)^2 = 4^2
Simplify:
(x+7)^2 + (y+6)^2 = 16
Answer:
Step-by-step explanation:
The graph shows the solution (-6,2)
i.e at x= -6 y=2
Analysis of each of the answers, since we can't write the equation of a straight line with only that information i.e the single point
Then,
Option 1
1. 2x - 3y = -6
x= -6 y=2
Then let insert x=-6 and y =2
2(-6)-3(2)
-12-6
-18.
Since -18 ≠ -6, then this is not the equation of the line and doesn't make up the system
Option 2
2. 4x - y = 26
Inserting x=-6 and y=2
4(-6)-(2)
-24-2
-26
Since -26 ≠ 26, then this is not the equation of the line and doesn't make up the system
Option 3
3. 3x + 2y = -14
Inserting x=-6 and y=2
3(-6)+2(2)
-18+4
-14
Since -14 ≠ -14 then this is the equation of the line and it make up the system.
Option 4
x-y = -2
Inserting x=-6 and y=2
(-6)-(2)
-6-2
-8
Since -8≠ -2, then this is not the equation of the line and doesn't make up the system
Option 5
5. x+y=-4
Inserting x=-6 and y=2
(-6)+(2)
-6+2
-4
Since -4 ≠ -4, then this is the equation of the line and it makes up the system.
Then, there are two option that make up the system
3. 3x + 2y = -14
And
5. x+y=-4
Pt 1 - how much the value goes down each year
Pt 2 - how much the car was originally worth
Axis of symmetry = -b/2a
y + 3x - 6 = -3(x-2)² + 4
y + 3x - 6 = -3 ( x²-4x+4) + 4
y + 3x - 6 = -3x² + 12x -12 + 4
y + 3x - 6 = -3x² + 12x -8
y = -3x² + 12x - 3x -8 + 6
y = -3x² + 9x -2
(a=-3, b=9, c=-2)
Axis of symmetry = -b/2a = -9/2(-3) = -9 / -6 = -3 / -2 = 3/2
Hence, option A) x=3/2 is the correct answer.
Answer:
The number of complex roots is 6.
Step-by-step explanation:
Descartes's rule of signs tells you that the number of positive real roots is 0. The number of negative real roots will be at most 2. The minimum value of the left side will be between x=0 and x=-1, but will never be negative. Thus all six roots are complex.
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The magnitude of x^3 will exceed the magnitude of x^6 only for values of x between -1 and 1. Since the magnitude of either of these terms will not be more than 1 in that range, the left-side expression must be positive everywhere.
