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2 years ago

Why is it important to learn hospitality law?

Advanced Placement (AP)

1 answer:

Volgvan2 years ago

3 0

Answer:

It is important for industry entrepreneurs and managers to know these laws to promote employee welfare, ensure adequate service to guests, manage business exposure to risk, and maintain complete compliance with state and federal laws. hope this helps :)

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What happened following the decision in brown v. board of education

Olin [163]

<span>Eight years after the U.S. Supreme Court ruled unanimously that racial segregation in public schools was unconstitutional, in the 1954 case of Brown v. Board of Education of Topeka, he penned this cartoon expressing his dismay at the country's slow progress toward educational integration.</span>

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2 years ago

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What’s the correct answer

dmitriy555 [2]

This is pretty dark lol, but I would say "gunshot wound" I have never learned this but it is my best guess.

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2 years ago

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In the statement "Critical thinking requires reasoned judgments," the word reasoned means

timama [110]

Answer:

In the statement "Critical thinking requires reasoned judgments," the word reasoned means logical and well thought out.

Explanation:

7 0

2 years ago

Hello, I need help with a calculus FRQ. My teacher has given a hint that this last part has to do with the intermediate value th

lesya [120]

Answer:

Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

Explanation:

To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

For Chloe's velocity

$C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2$

Finding the boundaries of the function gives;

$0\times e^{4-0^2} = 0$ and $2\times e^{4-2^2} = 2$

At t = 1, we have $1\times e^{4-1^2} = e^{3} = 20.086$

We find the maximum point as follows;

$\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0$

From which we have;

$\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0$

$e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0$

$e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\$

$e^{4-t^2}=0$ or $(1 - 2\times t^2 })=0$

∴ 1 = 2·t² and from which t = (√2)/2

Hence the function C(x) is decreasing from t = (√2)/2 to t = 2

For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

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2 years ago

Draw a supply and demand graph and label the equilibrium price as P and equilibrium quantity as Q. Then show what would happen t

Pavel [41]

Answer:

omg,thank you so much,you arw so good

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2 years ago