Answer:
The absolute maximum is
and the absolute minimum value is 
Step-by-step explanation:
Differentiate of
both sides w.r.t.
,


Now take 



![\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]](https://tex.z-dn.net/?f=%5CRightarrow%201-2%5Csin%20%5E2t%20%3D%5Csin%20t%20%20%5Cquad%20%5Cquad%20%20%5B%5Cbecause%20%5Ccos%202t%20%3D%201-2%5Csin%20%5E2t%5D)






In the interval
, the answer to this problem is 
Now find the second derivative of
w.r.t.
,

![\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%5Bf%27%27%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D-2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2-4%5Ctimes%20%5Cfrac%7B%5Csqrt%203%7D2%3D-3%5Csqrt%203)
Thus,
is maximum at
and minimum at 
![\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0](https://tex.z-dn.net/?f=%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2%2B%5Cfrac%7B%5Csqrt%203%7D2%3D%5Cfrac%7B3%5Csqrt%203%7D2%5C%3B%5Ctext%7Band%7D%5C%3B%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%7B%5Cpi%7D2%7D%3D%202%5Ctimes%200%2B0%3D0)
Hence, the absolute maximum is
and the absolute minimum value is
.
Answer:
1.) Amount of potting soil purchased last week.
2) 2p + 4 = 28
3.) 12 lbs
Step-by-step explanation:
1.) The unknown information is what we intend to find out using the information given, which is the amount of potting soil Emmanuel bought last week.
2.)
Let amount of potting soil bought last week = p
Amount bought this week = 2p + 4
Since amount bought this week = 28
Hence, we have ;
28 = 2p + 4
28 - 4 = 2p
24 = 2p
p = 24 / 2
p = 12
Answer:
-3
Step-by-step explanation:
2x-1=4x+5
-1=2x+5
-6=2x
x=-3