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skad [1K]
3 years ago
13

Can someone tell me the right answer to this question it is algebra plss and thank you

Mathematics
1 answer:
fgiga [73]3 years ago
6 0

b the table represents a genetic

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Make k the subject of 7k=4k/3t-11t
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Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].
castortr0y [4]

Answer:

The absolute maximum is \frac{3\sqrt 3}2 and the absolute minimum value is 0.

Step-by-step explanation:

Differentiate of f both sides w.r.t.  t,

f(t)=2 \cos t+\sin 2t

\Rightarrow f'(t)=-2\sin t+2\cos 2t

Now take f'(t)=0

\Rightarrow -2\sin t+2\cos 2t=0

\Rightarrow 2\cos 2t=2\sin t

\Rightarrow \cos 2t=\sin t

\Rightarrow 1-2\sin ^2t =\sin t  \quad \quad  [\because \cos 2t = 1-2\sin ^2t]

\Rightarrow 2\sin ^2t+\sin t-1=0

\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0

\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0

\Rightarrow (\sin t+1)(2\sin t-1)=0

\Rightarrow \sin t+1=0  \;\text{and}\; 2\sin t-1=0

\Rightarrow \sin t =-1  \;\text{and}\;   \sin t =\frac 12

In the interval 0\leq t\leq \frac {\pi}2, the answer to this problem is \frac {\pi}6

Now find the second derivative of f(t) w.r.t.   t,

f''(t)=-2\cos t-4\sin 2t

\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3

Thus, f(t) is maximum at t=\frac {\pi}6 and minimum at t=0

\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

Hence, the absolute maximum is \frac{3\sqrt 3}2 and the absolute minimum value is 0.

7 0
3 years ago
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vampirchik [111]
\frac { 7 }{ 28 }

28 is 4 times 7, so we will rewrite it like that.

28=4\cdot 7\\ \\ \frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 }

7 is 7 times 7, so we will rewrite it like that.

\frac { 7 }{ 28 } =\frac { 7 }{ 4\cdot 7 } =\frac { 1\cdot 7 }{ 4\cdot 7 } =\frac { 1 }{ 4 } \cdot \frac { 7 }{ 7 } =\frac { 1 }{ 4 } \cdot 1=\frac { 1 }{ 4 }

So the simplest form is,

\boxed { \frac { 1 }{ 4 }  }
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