Answer:
b is the answer
Step-by-step explanation:
I took the test
Easy
y=a(x-h)^2+k
vertex is (h,k)
we know that vertex is (4,0)
input that point for (h,k)
y=a(x-4)^2+0
y=a(x-4)^2
passes thorugh the point (6,1)
input that point to find a
1=a(6-4)^2
1=a(2)^2
1=a(4)
divide both sides by 4
1/4=a
thefor the equation is
y=(1/4)(x-4)^2
or
y=(1/4)x^2-2x+4
Answer: 105 miles?
I am not sure but that"s what I got
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
(a)
OC = OB ( both radii of the circle )
Thus Δ BOC is isosceles with congruent base angles.
∠ BOC = ∠ BCO = 50°
(b)
∠ ACB = 90° ( angle in a semicircle ), then
∠ ACO = 90° - 50° = 40°
OA = OC ( both radii of the circle )
Thus Δ ACO is isosceles with congruent base angles.
∠ BAC = ∠ ACO = 40°
Answer:
All real numbers greater than or equal to -3
Step-by-step explanation:
Notice for which values of the x-axis the function gives a well defined y-value (indicated by the trace of the curve in blue).
There is a solid dot at the point -3 for x and -2 for y, where the trace of the function begins. That means that the function is defined by f(-3) = -2.
and all x values to the right of -3 seem to also have a well defined y value f(x) that is represented by the blue curve.
Therefore, all x values starting at -3 (including it) and to the right (larger than -3) have well defined associated y values. Such constitutes the actual Domain of f(x).
