B) 9(a+b)+4 (3a+2b)=21a+17b
Given:
The equation of the curve is:

To find:
The gradient of the given curve at the point where
.
Solution:
We have,

Differentiate with respect to x.
![[\because \dfrac{d}{dx}x^n=nx^{n-1}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cdfrac%7Bd%7D%7Bdx%7Dx%5En%3Dnx%5E%7Bn-1%7D%5D)

Substituting
, we get


Using properties of exponents, we get
![[\because (a^m)^n=a^{mn}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a%5Em%29%5En%3Da%5E%7Bmn%7D%5D)



Therefore, the gradient of the given curve at the point where
is 20.
Multiply 8/10 by 6 and you get:
48/60
Noah’s heartbeat is 48 beats per minute. Well, then, Noah must be Usain Bolt...
Noah’s heartbeat is not typical.
Answer:
Oook, quick setup: pick a cartesian plane, be
the position of the helicopter as time passes, and
your position as you start running. In my horrible sketch, the green line is the distance when you start running, the red line is the distance you need, in general what you want is to move one end of the segment up, the other right, and stretch it.
That distance is easily found with the formula for the distance between two points in the plane, or
(playing a bit with the squares). At this point we need to write down expressions for both
and find at what value of t we need to evaluate our distance.
For the sake of semplicity, let's start measuring times the moment you start running. Since speed is constant, both expressions will be of the form 
The equation for the runner becomes simply
, since you haven't moved until the heli is high enough.
The equation of the helicopter is
since the heli is 30ft above ground when you start running.
Finally, we need to know how many seconds have passed when the heli is at 60 ft above ground. that happens when
. in this time, you are at
from the origin.
Plugging it in the distance formula you get: 