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2)

\(f(x) = \dfrac{3}{9^x + 3}\\ f(1-x) = \dfrac{3}{9^{1-x}+3}\\ f(1-x) = \dfrac{3\cdot 9^x}{9+3\cdot 9^x}\\ f(1-x) = \dfrac{9^x}{3+9^x}\\ \therefore f(x) + f(1-x) = 1 \quad\forall x\in \mathbb R\)

\(\quad\text{Answer}\\=\displaystyle \sum_{k=1}^{1000}f\left(\dfrac{k}{1001}\right)\\ = \displaystyle \sum_{k=1}^{500} \left(f\left(\dfrac{k}{1001}\right) + f\left(1-\dfrac{k}{1001}\right)\right)\\ = \displaystyle \sum_{k=1}^{500} 1\\ = 500\)

No! I’m not saying that. Your hypothetical is inconsistent with the data points in the question.

My critique focuses on the meaning of the word “above”as used in the context of this question.

The solution equation for “of” is (1600 x 3)-800 = 4000 dollars.

The solution equation for “above” is (1600 x 4)-800 =5600 dollars. This should be the correct solution. The word “above” is in the question. The word “of” is not.  This is what guest was attempting to demonstrate when he asked the pointed question.  It should be clear that $1600 is not above $1600.  This being true means the solution included with the origional question is wrong.

GA

😧 what an extremely long critique. So you are saying that the store could, (hypothetically) bring in 10000000000000000000000000000000000000000000000000000 dollars suddenly?

since it could be above average?

I have never thought about it like that.

1) 

Compute the product

\(\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}} \\\\ &=& \dfrac{(1998 - 1996)(1998 + 1996)(1998 - 1995)(1998 + 1995) \dotsm (1998 - 0)(1998 + 0)} {(1997 - 1996)(1997 + 1996)(1997 - 1995)(1997 + 1995) \dotsm (1997 - 0)(1997 + 0)} \\\\ &=& \dfrac{2\cdot 3994 \cdot 3 \cdot 3993 \dotsm 1998\cdot 1998 } {1\cdot 3993 \cdot 2 \cdot 3992 \dotsm 1997\cdot 1997 } \\\\ &=& \dfrac{2\cdot 3 \dotsm 1998 } {1 \cdot 2 \dotsm 1997 } \times \dfrac{3994 \cdot 3993 \dotsm 1998 } {3993 \cdot 3992 \dotsm 1997 } \\\\ &=& 1998 \times \dfrac{3994 } {1997 } \\\\ &=& 1998 \times \dfrac{2\cdot 1997} {1997 } \\\\ &=& 2\cdot 1998 \\ &=& \mathbf{3996 } \\ \hline \end{array}\)

This question, as presented, has two (2) syntax errors:

During a holiday month a retail store brings in 300% above it's average sales in other months. If a typical month has $1600 in sales, and has fixed cost of $800 per month. What is the profit for a holiday month? 

... A missing comma, and a wrong word (“it’s”) in the first sentence, and a sentence fragment in the second sentence.

Correcting these errors, the sentence reads:

During a holiday month, a retail store brings in 300% above its average sales in other months. If a typical month has $1600 in sales, and has fixed cost of $800 per month, what is the profit for a holiday month? 

Word problems often have ambiguity in them, but this question is clear (to me).

As the question is asked, the solution presented (1600 x 3)-800=4,000 is wrong. The reason it’s wrong is the word “above” is in the phrase “a retail store brings in 300% above its average sales in other months.”

If the quest read:

“During a holiday month, a retail store brings in 300% of its average sales in other months.” Then the asker’s solution would be correct and so would your answer to guest’s question.  This was the point of guest’s question.

Your answer to guest’s question clearly indicates the result is equal to the average, not above the average as indicated in the question. (Note that this is the earnings, not the profit as requested by the question.)

Wow, Thank you for a very detailed answer to a mistake in my answer.

The 50% is indicating the probability I actually get it right.

Sincerely,

\(tommarvoloriddle\),

Sorry about the previous answer. A family member trolled me.

I have done these types of problems before, just that i don't get it right often, and I consider that a never ever done it before. Does that help?

Tommarvoloriddle, where does the 50% come from? Usually when you use a numeric value indicating a probability, there should be a mathematical basis for it.  This is doable for non-tangible concepts, such as feelings, but this requires evaluative experience with similar problems.   

For example, if you have previously evaluated (N) of these types of problems, and the tabulations of the accuracy your assessments indicate that you are correct 50% of the time, then your 50% certainty is valid.  However, your final statement says you have “never done these types of problems before ....”  Knowing this, your confidence should be very close to zero (0) instead of the 50% you indicated. 

Over time, intuition (feelings) will form as you continue to develop your skills in mathematics. Intuition is a very common thought process in all skilled, Master and PhD-Level mathematicians and scientists. However, such intuition is never an acceptable substitute for a demonstrative proof of concept.      

Offering opinions without the minimum required skill sets in the subject puts you in to the BS class. The world is full of amateur and professional BSers in and on every subject.  We do not need anymore! Until you develop these skills, you should keep your comments at inquiring observer level.  

Find a 2*2 matrix A such that       \(\binom{2 \text{ } 7}{5\text{ } 17}A= \binom{3 \text{ } -2}{4 \text{ } 1}\)

\(\binom{2 \text{ } 7}{5\text{ } 17}^{-1}=\frac{1}{2*17-7*5}\binom{17 \text{ } -7}{-5\text{ } 2}=\binom{-17 \text{ } 7}{5\text{ } -2}\)

\(A=\binom{-17 \text{ } 7}{5\text{ } -2}\binom{3 \text{ } -2}{4\text{ } 1}=\binom{-23 \text{ } 41}{7\text{ } -12}\)

No problem Dream driven!

Thanks so much for your honesty and input.

I think they would just earn 1600... not sure.

ok. I feel that your answer is right... Like I think it has a 50% of being right. I have never done these types of problems before, so I am not sure.

I could be wrong but I think it may be the yearly maintenance cost to do maintenance on a product or upgrades on a certain piece of software or hardware?

only that what.

Quick question: what is annual maintenance?

Thanks,

\(tommarvoloriddle\)

a + ab²  =  40b

a - ab²  =  -32b

The purple values are equal and the blue values are equal.  purple + blue  =  purple + blue

(a + ab²) + (a - ab²)  =  40b + -32b

(a + ab²) + (a - ab²)  =  40b + -32b

2a  =  8b

\(\frac14\)a  =  b

Now we can substitute this value for  b  into one of the original equations.

a + ab²  =  40b

                                    Substitute   \(\frac14\)a   in for   b

a + a(\(\frac14\)a)²  =  40(\(\frac14\)a)

                                    Simplify both sides of the equation.

a + \(\frac{1}{16}\)a³   =   10a

                                    Multiply through by  16

16a + a³  =  160a

                                    Subtract  16a  and subtract  a³  from both sides

0  =  144a - a³

                                    Factor  a  out of both terms on the right side

0  =  a( 144 - a² )

                                           Factor   144 - a²   as a difference of squares

0  =  a( 12 - a )( 12 + a )

                                           Set each factor equal to  0  and solve for  a

this question does not have an answer on AoPS. Can someone please help me, I am stuck.

You should upload the actual picture.

No "AoPS" questions on this forum!

yes..its something like that only that