Please, use " ^ " to denote exponentiation: p(t) = t^2 + 5t + 6.
To find the critical points, differentiate p(t) with respect to t, set the result = to 0, and then solve the resulting equation for t:
p '(t) = 2t + 5 = 0
Solving for t: 2t = -5, and so t = -5/2. (-5/2, p(-5/2)) is the critical point. That evaluates to (-5/2, -0.25). This happens to be the vertex of a parabola that opens up.
Answer:
x-5=-3
Step-by-step explanation:
|x-5|+2=5
|x-5|=5-2
|x-5|=3
x-5=±3
so x-5=-3 is a correct step.
73℅ of 58
= (73/100) × 58
= (73×58)/100
= 4234/100
= 42.34
<span>4264.00 this is the answer
hope it helps
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