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-Dominant- [34]

2 years ago

9

A few students are having a bake sale to try to raise more than $600 for a class trip. They sell cupcakes (c) for $2 each and pi

es (p) for $5 each. If the students have already sold at least 30 cupcakes, which of the following systems of inequalities represents the situation?

2 answers:

Alenkinab [10]2 years ago

6 0

Thats easy its D hope this helps

nadya68 [22]2 years ago

4 0

Answer:

it is easy the answer is D.

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e-lub [12.9K]

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I will make the assumption that <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10.
</span>
$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use <span>Lhopitals rule

</span> $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$
<span>
Therefore,

</span> $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$ <span>
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