Siny/2.7=sin63/2.8
siny=2.7sin63/2.8
y=arcsin((2.7sin63)/2.8)
y≈59.23° (to nearest one-hundredth of a degree)
A turning point occurs when the velocity is equal to zero, but the acceleration is not equal to zero.
t(x)=(x+5)^3+7
dt/dx=3(x+5)^2
d2t/dx2=6(x-5)
dt/dx=0 only when x=-5
However, since d2t/dx2(-5)=0, this point is an inflection point, not a turning point.
So there is no turning point for this function.
Now in this problem, it is even easier than the above to show that there is no turning point. A turning point by definition is when the derivative or velocity changes sign. Since in this case v=3(x+5)^2, for any value of x, v≥0, and thus never becomes negative, so it never changes from a positive to negative velocity because velocity in this instance is a squared function.
Answer:
As soon as the car travels south you would take 5 miles off the 8 south which would make it 3 miles but you would have to add an extra 6 as it also travelled west so the car would have to travel east to arrive at the initial starting point.
Step-by-step explanation:
I hope this helps you
q. (3+t)=2-4t
q.3+q.t=2-4t
4t+q.t=2-3q
t (4+q)=2-3q
t=2-3q/4+q
