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PIT_PIT [208]
4 years ago
6

- Suppose Q is the midpoint of PR. Use the information to find the missing value. PQ = 3x + 14 and QR = 7x - 10; Find x.​

Mathematics
1 answer:
azamat4 years ago
7 0

Answer:

x =6

Step-by-step explanation:  Q is the midpoint of PR.

Therefore PQ = QR

3 x + 14 = 7 x -10

       -14          -14

3 x         = 7 x-24

-7x                 -7x

-4x     =      -24                             Divide both sides by -4

x = 6

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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]

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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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