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Alchen [17]
3 years ago
8

Subtract 1/4w+4 from 1/2w+3 .

Mathematics
2 answers:
dezoksy [38]3 years ago
4 0
I think the correct answer is

- w = - 1
w = 1
True [87]3 years ago
3 0

Answer:  1/4w - 1


Step-by-step explanation:  just had a test

hope this helps



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When does the net area of a region equal the area of a​ region? when does the net area of a region differ from the area of a​ re
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Before answering this question let us know what are net and total areas.

\text{Net area}= \text{total area above x-axis}-\text{total area below x-axis}

\text{Total area}= |\text{total area above x-axis}|+|\text{total area below x-axis}|

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Calculate the limit of the function with L'Hospital rule​
mr_godi [17]

Answer:

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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

6 0
3 years ago
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