Answer:
y = -1/4x + 3
Step-by-step explanation:
Because you are finding the perpendicular slope, you need to find the negative reciprocal of the original line, which would be -1/4. You then use point slope form to find the y-intercept with the slope and given point:
y - 5 = -1/4(x + 8). That equals to y = -1/4x + 3.
So the equation of this line is y = -1/4x + 3.
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.
We went from -12 degrees to 27 degrees.
27 - (-12) = 27 + 12 = 39
Answer:
Correct answer: sin x ⇒D(x) : [- π/2, π/2] ; sin⁻¹x ⇒ CD(x) : [- π/2, π/2]
Step-by-step explanation:
In order for the function sin x to have an inverse function sin⁻¹x due to the monotony, the domain is taken D(x) : [- π/2, π/2] and the range of sin⁻¹x is CD(x) : [- π/2, π/2].
God is with you!!!
