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Hunter-Best [27]
2 years ago
9

Which compound inequality is equivalent to | ax -b | > C for all real numbers a, b, and c, where c>0?

Mathematics
1 answer:
My name is Ann [436]2 years ago
8 0

Hello,

Answer D

ax-b < -c or ax-b >c

If ax-b >0 then |ax-b| = ax-b ==> ax-b > c

if ax-b <0 then |ax-b|=-(ax-b) > c ==> ax-b <-c

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A runner sprints around a circular track of radius 100 m at a constant speed of 7 mys. The runner’s friend is standing at a di
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\frac{7\sqrt{15}}{4}\text{ meter per sec}

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Let B represents the position of runner, A represents the position of the friend and C represents the position of centre of the circular track. ( shown  below ),

We need to find : \frac{dc}{dt}

By the cosine law,

c^2 = a^2 + b^2 - 2ab \cos C

Differentiating with respect to t ( time ),

2c\frac{dc}{dt} = 2ab \sin C \frac{dC}{dt}

\implies \frac{dc}{dt}=\frac{ab \sin C\frac{dC}{dt}}{c}-----(1)

Now, by arc length formula,

Radius( say r ) × angle = arc length ( say l )

r\times \angle C = l

Differentiating w. r. t. t,

r\times \frac{dC}{dt}+\angle C\times \frac{dr}{dt}= \frac{dl}{dt}

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200^2 = 100^2 + 200^2 - 2(100)(200)\cos C

40000 = 10000 + 40000 - 40000\cos C

-10000 = -40000\cos C

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\sin C = \sqrt{1-\cos^2 C}=\sqrt{1-\frac{1}{16}}=\sqrt{\frac{16-1}{16}}=\frac{\sqrt{15}}{4}---(3)

From equation (1), (2) and (3),

\frac{dc}{dt}=\frac{(100)(200)\frac{\sqrt{15}}{4}\frac{7}{100}}{200}=\frac{7\sqrt{15}}{4}\text{ meter per sec}

Hence, the distance between the friends changing with the rate of \frac{7\sqrt{15}}{4} meter per sec.

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