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seropon [69]
3 years ago
10

Maya bought 8 chicken wings for $8.80. If Maya spent $25.30, how many chicken wings did she buy?

Mathematics
1 answer:
kupik [55]3 years ago
6 0

Answer:

8 wings / $8.80 = $1.10 per chicken wing

$25.30 / 1.10 = 23 chicken wings

Step-by-step explanation:

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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
FREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POI
monitta

Answer:

OMG THANK YOU SO MUCH I NEEDED THIS SO MUCH FOR MY MATH TEST!!!!

Step-by-step explanation:

7 0
2 years ago
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