What is 130% of 56g?

Mathematics

1 answer:

jeka57 [31]3 years ago

5 0

The answer would be 72.8 grams

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$\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}$

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$\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)$

Then using the known limit above, it follows that

$\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}$