Answer:
f(5) = 17
Step-by-step explanation:
Pretty easy :)
just pluggin f(5).
f(5) = 3(5) + 2
f(5) = 15+2
f(5) = 17.
:)
It's 3. 3 can divide into all those numbers evenly and be the one with the smallest value
You’re solving for x, so the first step is to move all the x’s to the same side. Subtract 2x from both sides. You’re left with:
2x + 5 = -3
The x variable still isn’t by itself. Subtract five from both sides.
2x = -8
The last step to get x by itself is to divide both sides by 2.
Your final answer is: x = -4
Answer:
1/q^30 (startfraction 1 over q subscript 30 baseline end fraction)
Step-by-step explanation:
just took quiz
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.