simplify the following polynomial expression.
B%28-2x%5E%7B3%7D%20%2B3x%5E%7B2%7D%20-7x%2B9%29-%285x%2B3%29%284x%5E%7B2%7D%20%2Bx-1%29" id="TexFormula1" title="(5x^{3} +4x^{2} +6x-3)+(-2x^{3} +3x^{2} -7x+9)-(5x+3)(4x^{2} +x-1)" alt="(5x^{3} +4x^{2} +6x-3)+(-2x^{3} +3x^{2} -7x+9)-(5x+3)(4x^{2} +x-1)" align="absmiddle" class="latex-formula">
1 answer:
You might be interested in
For this case we have that by definition, the slope of a line is given by:
Where:
and
are two points through which the line passes.
According to the figure we have to, the line goes through:
Substituting we have:
Thus, the slope of the line is -1.
Answer:
Answer:
Step-by-step explanation:
If the position function is
and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:
and the easiest way to do this is to factor by taking the GCF of -40t:
0 = -40t(t - 5) and by the Zero Product Property,
-40t = 0 or t - 5 = 0. Solving for t, we get
t = 0 (which is before the object is launched) and
t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).
Your choice is c) 5