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Galina-37 [17]
4 years ago
11

If you have 60 and get 45 as discount how much the percentage

Mathematics
1 answer:
WINSTONCH [101]4 years ago
7 0
Let x% be the percentage
60×(1-x%)=45
1-x%=0.75
x=25%
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There are 16 boys and 20 girls in a class. What part of the class is boys? What part girls?
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Calculate the limit of the function with L'Hospital rule​
mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if \lim_{x \to a} \frac{f(x)}{g(x)} takes \frac{0}{0} or \frac{\infty}{\infty} form, then,

\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}

where f'(x)=\frac{df(x)}{dx} and g'(x)=\frac{dg(x)}{dx}

Now coming to the problem,

L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}

Here f(x)=cot^{3}x-3cotx and g(x)=cos(x+\frac{\pi}{3} )

Substituting x=\frac{\pi}{6} in f(x) and g(x),

f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0

g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0

Since L takes the form \frac{0}{0}, using l'hopital's rule

L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}

now substituting x=\frac{\pi}{6} ,

L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24

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The answer is

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