This is how you would graph y=5x+1
They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east. Then we can say that there positions, y and x are:
y=150-600t x=200-800t
By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:
d^2=y^2+x^2, using y and x from above
d^2=(150-600t)^2+(200-800t)^2
d^2=22500-180000t+360000t^2+40000-320000t+640000t^2
d^2=1000000t^2-500000t+62500
d=√(1000000t^2-500000t+6250)
So the rate of change is the derivative of d
dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)
dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)
So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...
...
To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...
1000000t^2-500000t+62500=0
6250(16t^2-8t+1)=0
6250(16^2-4t-4t+1)=0
6250(4t(4t-1)-1(4t-1))=0
6250(4t-1)(4t-1)=0
6250(4t-1)^2=0
4t-1=0
4t=1
t=1/4 hr
Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)
Answer:
Step-by-step explanation:
The slope of negative one means it is the opposite of a positive graph and the overall value is decreasing. The origin is (0,0)
A^2+b^2=c^2 so c^2=12^2+16^2 which then simplifies to c^2=144+256 then simplify that to c^2=400. After that take the square root of c^2 and the square root of 400. So your answer is c=20
Answer:
e^(ln x) is just plain x
Step-by-step explanation:
The functions f(x) = e^x and g(x) = ln x are inverses of one another. In other words, one "undoes" the other.
Thus, as the rule goes, e^(ln x) is just plain x.
Here, e^(ln x) = 4 simplifies to x = 4.
