Answer:
A y=mx+b y=3x+50
because $3 is the rate and $50 is what you get started off with
5+4v-v=1+3+5v
5+3v=4+5v
minus 3v both sides
5+3v-3v=4+5v-3v
5+0=4+2v
5=4+2v
minus 4 both sides
5-4=4-4+2v
1=0+2v
1=2v
divide both sides by 2
1/2=v
Answer:D
Step-by-step explanation:
Radius=r=8
Φ=2π/3
π=180
2π=360
length of arc=Φ/360 x 2 x π x r
Length of arc=(2π/3 ➗ 2π) x 2 x π x 8
Length of arc=(2π/3 x 1/2π) x 2 x π x 8
Length of arc=((2πx1)/(3x2π)) x 2 x π x 8
Length of arc=2π/6π x 2 x π x 8
Length of arc=1/3 x 2 x π x 8
Length of arc=(1 x 2 x π x 8)/3
Length of arc=16π/3
LxW 5x5=25 If you know one side is five than the other side is 52 because they’re both equal its the radius compared to the diameter radius is five the diameter is 10
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
