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.
It is 166 just do some side math if you can and I think youtline get it ok but that's what I got if someone eles gets somthing eles matey listen to them
That will depend, you have to ask yourself first how many kilobytes one picture is. Let's just say that the size of one picture is 100 kb (which is the average size of a picture) Then first you multiply 100 kb with the number of pictures which is 43. Now you have a total used up memory of 4300 kb. After that, you minus the used up memory which is 4300 kb, to the total available space which is 32,834.5 and you will get an available space of 28534.5 kb. After that, you divide the remaining available space with the size of each picture. So this will be 28534.5 divided by 100. You will get 285. You can still take 285 pictures.
Answer:
A, B, C
Step-by-step explanation:
Look at how many field goals were actually made and order them from least to greatest- 12, 15, 19. They are already listed from least to greatest in the table.
Answer: 2 197/1000
Explanation:
(1 3/10)^3
= (13/10)^3
= 13^3/10^3
= 2197/1000
= 2 197/1000
