Answer:
The answer to your proble I'm slow intercept from is y=-2x+8
Step-by-step explanation:
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:
3(x - 5)(x + 1).
Step-by-step explanation:
3x^2 - 12x - 15
Dividing through by 3:
= 3(x^2 - 4x - 5)
We need 2 numbers whose sum is -4 and whose product = -5. That is -5 and +1 , so the factors are:
3(x - 5)(x + 1).
Answer:
81
Step-by-step explanation:
just do it step by step adding them together.
