91.5 is the perimeter. It's just 5a.
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:
= -5
Step-by-step explanation:
1-6=5
1-6
{Subtract the numbers:} 1-6=-5
= -5
Since h represents the height of the ball at any given time, t, let h = 25, such that the ball will be 25m high at t.
Now, we have 25 = 20t - 5t²
5t² - 20t + 25 = 0
t² - 4t + 5 = 0
Since the discriminant is less than zero, there are no solutions.
Hence, the ball will never be 25m high.
Answer:
D
You use algebra and solve it for y.
Step-by-step explanation:
4x-3y=18
3y=4x+18
y=4/3x +6
