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Semmy [17]
4 years ago
7

HELP ILL GIVE A BRAINLIEST NEED DONE SUPER QUICKLIY!!!!!

Mathematics
2 answers:
Free_Kalibri [48]4 years ago
7 0

slope = -50

 Amount of dirt removed per load

900

total amount of dirt to remove


eduard4 years ago
3 0

Answer:

1. The numerical value of the slope is:

Lets take 2 coordinates.

(14,200)  (16,100)

Slope is \frac{y_2-y_1}{x_2-x_1}

\frac{100-200}{16-14}

= \frac{-100}{2}

= -50

2. The slope represents: Amount of dirt removed per load  (that is the level of the dirt is decreasing with each load)

3. The numerical value of the y-intercept is : 900

The y intercept's scale denotes 50 for each point.

The y intercept is 3 points above 750 means 3\times50=150 added to 750 gives 900.

4. The y-intercept represents the volume of dirt that needs to be removed. So, answer is :The total amount of dirt to remove.

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d, e, g, and h.

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3 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.
7 0
3 years ago
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