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3 years ago

Applying L'Hospital's rule is not allowed

Mathematics

1 answer:

Ivan3 years ago

3 0

Answer:

-1/2

Step-by-step explanation:

lim x-> π/2 cos x /(2x-π) =

lim (x-π/2)->0 sin (π/2 - x) /2(x-π/2) =

lim (x-π/2)->0 - sin (x - π/2)/2(x-π/2) = -1/2

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A dollar bill is worth _____ pennies. When written in decimal format, each penny can be written as ______ of a dollar. 100:10 10

erik [133]

Hey there! :D

A dollar bill is worth 100 pennies.

$1.00-> .01*100 <=== same thing

When written in decimal format, each penny can be written as 100:1, because there are 100 pennies in a dollar. One penny would just be one out of 100.

I hope this helps!
~kaikers

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3 years ago

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A crater shaped like a cone has a height of 24 yd and a diameter of 72 yd.

mezya [45]

The formula for the volume of a cone is (1/3)*area of the base * height. The base is a circle, whose area is pi times radius^2. Here this is pi * (r^2) = 3.14*(72 / 2 yd)^2 = 4069.44 yd^2. And so the volume is (1/3) * 4069.44 yd^2 * 24yd = 32,555.52 yd^3. I recommend that you find the most common formulas of the geometrical figures and make a flashcard that you can use at any moment.

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3 years ago

Math Plz Help. Please graph with two points! explain how you found your answer

Softa [21]

X intercept: y = 0
y= -3/4x -4
0 = -3/4 x - 4
4= -3/4 x
4 • -4/3 = x
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4 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

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.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

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

So in the limit you get

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

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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Radda [10]

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