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Ivanshal [37]
3 years ago
10

Consider a circle with a diameter of 10cm what is the circumference of the circle. What is the ratio of the circumference to the

diameter
Mathematics
1 answer:
Nutka1998 [239]3 years ago
4 0

Answer:

Circumference formula: A = π * diameter

3.14×10=31.4

31.4:10

or 3.14 to 1

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Derivative of tan(2x+3) using first principle
kodGreya [7K]
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The derivative is given by the limit

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Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

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\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

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\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
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