Answer:
In words the answer is between t=0 and t=2.
In interval notation the answer is (0,2)
In inequality notation the answer is 0<t<2
Big note: You should make sure the function I use what you meant.
Step-by-step explanation:
I hope the function is h(t)=-16t^2+32t because that is how I'm going to interpret it.
So if we can find when the ball is on the ground or has hit the ground (this is when h=0) then we can find when it is in the air which is between those 2 numbers.
0=-16t^2+32t
0=-16t(t-2)
So at t=0 and t=2
So the ball is in the air between t=0 and t=2
Interval notation (0,2)
Inequality notation 0<t<2
It's hard to type and hard to read the "inverse tangent" function, as you've seen (above).
So, use "arctan x" instead.
Then the problem becomes: "differentiate cos (arctan x)."
You must apply first the rule for differentiating the cosine function, and next apply the rule for differentiating the arctan function:
(d/dx) cos (arctan x) = - sin (arctan x) * [1/(1+x^2)]
Answer:
C = 132°
Step-by-step explanation:
(5y + 28) = 12y
5y - 5y + 28 = 12y - 5y
28 = 7y
y = 4
12 (4)
= 48
48 + C = 180°
48 - 48 + C = 180 - 48
C = 132°
Answer:
- 11
Step-by-step explanation:
- 8 - 5 + 2
- 13 + 2
- 11
Answer:
2axay - 2bxby is the answer