Answer:
you determine net force of an object by using this formula:
F=M×A or force=mass×acceleration
really hope this helps:)
Use a=
r^2 (r represents your radius of 4 feet)
(4^2)
Answer = 16
^2
a = 30
a = 16
-a² + 46a -480 = 0
(-a + 30) (a -16) = 0
(-a)(a) + (-a)(-16) + 30(a) + 30(-16) = 0
-a² + 16a + 30a -480 = 0
-a² + 46a - 480 = 0
(-a + 30) = 0 ; (a - 16) = 0
-a = -30 ; a = 16
a = 30
To check:
a = 30
-(30)² + 46(30) - 480 = 0
-900 + 1380 - 480 = 0
480 - 480 = 0
0 = 0
a = 16
-(16)² + 46(16) - 480 = 0
-256 + 736 - 480 = 0
480 - 480 = 0
0 = 0
Answer:
y = 2x + 2
Step-by-step explanation:
The general form
is;
y = mx + b
where m is the slope and b is the y-intercept
so we have;
y = mx + 2
To get m, substitute the point (-1,0)
So we have
0 = -m + 2
m = 2
So the equation is;
y = 2x + 2
Answer:
<u>x-intercept</u>
The point at which the curve <u>crosses the x-axis</u>, so when y = 0.
From inspection of the graph, the curve appears to cross the x-axis when x = -4, so the x-intercept is (-4, 0)
<u>y-intercept</u>
The point at which the curve <u>crosses the y-axis</u>, so when x = 0.
From inspection of the graph, the curve appears to cross the y-axis when y = -1, so the y-intercept is (0, -1)
<u>Asymptote</u>
A line which the curve gets <u>infinitely close</u> to, but <u>never touches</u>.
From inspection of the graph, the curve appears to get infinitely close to but never touches the vertical line at x = -5, so the vertical asymptote is x = -5
(Please note: we cannot be sure that there is a horizontal asymptote at y = -2 without knowing the equation of the graph, or seeing a larger portion of the graph).
