Answer:
tehe
Step-by-step explanation:
tehetehetehetehe
Hey there,
Your question states: <span>Four points are always coplanar if . . .
Your correct answer from the questions above would be
</span>
they lie in the same place
The definition of the word
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means : In the same place.
So . .Four points are always coplanar if <span>
they lie in the same place.Hope this helps many.
~Jurgen</span>
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:
The standard form of the quadratic equation is x² + 3·x - 4 = 0
Step-by-step explanation:
The standard form of a quadratic equation is a·x² + b·x + c = 0
Given that the expression of the quadratic equation is (x + 4)·(x - 1) = y, we can write the given expression in standard form by expanding, and equating the result to zero as follows;
(x + 4)·(x - 1) = x² - x + 4·x - 4 = x² + 3·x - 4 = 0
The standard form of the quadratic equation is x² + 3·x - 4 = 0
The graph of the equation created with MS Excel is attached
The line passing through it is y=7/4x-11/<span>4</span>