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:
Given that, Vasudevan invested ₹ 60,000
For Compound Interest (C.I.)
A = P[1 + (r/100)]n
P = ₹ 60,000
n = 6 months and 1 year
R = 12% p.a. compounded half-yearly
where , A = Amount, P = Principal, n = Time period and R = Rate percent
(i) For easy calculation of compound interest, we will put Interest Rate as 6% half-yearly and n = 1.
Compound Interest to be paid for 6 months
A = P[1 + (r/100)]n
A = 60000[1 + (6/100)]1
A = 60000[(100/100) + (6/100)]
A = 60000 × (106/100)
A = 60000 × 1.06
A = ₹ 63600
(ii) Compound Interest to be paid for 12 months (1 year) compounded half yearly.
So, assume n = 2, r = 6%
A = P[1 + (r/100)]n
A = 60000[1 + (6/100)]2
A = 60000[(100/100) + (6/100)]2
A = 60000 × (106/100) × (106/100)
A = 60000 × (11236/10000)
A = 60000 × 1.1236
A = ₹ 67416
Answer:
41 quarts (or 10.25 gal)
Step-by-step explanation:
1 gal = 4 quarts
5 gal = 4 x 5 = 20 quarts
4 gal = 4 x 4 = 16 quarts
hence 4 gal & 3 qts = 16 + 3 = 19 quarts
5 gal + 2 qts + 4gal 3 qts
= 20 quarts + 2 quarts + 19 quarts
= 41 quarts ( = 41/4 = 10.25 gal)
School A : 12/45 = 0.266 = 26.6%
school B : 15/40 = 0.375 = 37.5%
school C : 9/27 = 0.333 = 33.3%
school D : 13/32 = 0.406 = 40.6%
so ur answer is school D
