Summarize the following:
Don't just walk on into a HC Party if you are a norm, and don't crash Norm parties if you're HC.

summary: Know your boundaries!


Summarize the following:
In calculus, the derivative of any function is used to find the rate of change of that function. The maximum value of a given function occurs when the derivative equals zero. So, to maximize the revenue, find the first derivative of the revenue function.  Suppose the revenue function, in terms of number of units sold, is R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}}. The first derivative, therefore, is: R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q}  For a review of derivatives, see the wikiHow article on how to Take Derivatives. When the derivative is zero, the graph of the original function is at either a peak or a trough. This will be either the maximum or minimum value. For some higher level functions, there may be more than one solution to the zero derivative, but not a basic price-demand function.  R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} 0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} Use basic algebra to solve the derivative for the number of items to sell where the derivative is equal to zero. This will give you the number of items that will maximize the revenue.  0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} 250q=500{\displaystyle {\frac {2}{50}}q=500} 150q=250{\displaystyle {\frac {1}{50}}q=250} q=50∗250{\displaystyle q=50*250} q=12,500{\displaystyle q=12,500} Using the optimal number of sales from the derivative calculation, you can enter that value into the original price formula to find the optimal price.  p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p=500−15012,500{\displaystyle p=500-{\frac {1}{50}}12,500} p=500−250{\displaystyle p=500-250} p=250{\displaystyle p=250} After you have found the optimal number of sales and the optimal price, multiply them to find the maximum revenue. Recall that R=p∗q{\displaystyle R=p*q}. The maximum revenue for this example, therefore, is:  R=p∗q{\displaystyle R=p*q} R=(250)(12,500){\displaystyle R=(250)(12,500)} R=3,125,000{\displaystyle R=3,125,000} Based on these calculations, the optimal number of units to sell is 12,500, at the optimal price of $250 each. This will result in a maximum revenue, for this sample problem, of $3,125,000.

summary: Find the first derivative of the revenue function. Set the derivative equal to 0. Solve for the number of items at the 0 value. Calculate the maximum price. Combine the results to calculate maximum revenue. Summarize the results.


Summarize the following:
More than in any other form of writing, diction and word choice matter in poetry. Try using descriptive words that paint a more elaborate picture. For example, you could say that it was a dark, shadowy night instead of just that the night was dark. This is much more descriptive and gives the reader a more accurate picture of what you mean. Metaphors directly compare two things based on similarities by equating them as the same. In his play As You Like It, William Shakespeare famously says, All the world's a stage, / And all the men and women merely players: / They have their exits and their entrances. This is a metaphor that compares the action of real life with the action of a theatrical play. Shakespeare says that world is a stage and all the people are actors, not merely that they are like actors. Analogies are comparisons between two things that intend to help the reader understand a situation or event. Usually, authors compare a known thing with a lesser known thing to help the reader understand the lesser known thing. Unlike metaphors that compare a thing by saying it is another thing, analogies say that something is like something else. For example, saying she was as quiet as a mouse is an analogy that lets the reader know something about the subject, she, by relating a fact about her to a fact everyone knows (that mice are quiet).
summary: Pay attention to your word choice. Implement metaphors. Use analogies.