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<p>I assume that you didn't know that you were supposed to multiply by -1. It probably means that you should do more inequality problems until the idea of multiplying by -1 becomes automatic.</p>

<p>Also, for p.400 I've figured it out, you don't even need a graph. I'm pretty sure that the shaded region is where a and b are both positive, right? In other words, in Quadrant I.</p>

<p>Ok. </p>

<p>There is a difference between a and f(a).</p>

<p>Let me give you an example:</p>

<p>f(x) = x + 2.</p>

<p>Now, you're probably aware that f(x) and x are not the same. The same goes with f(a) and a. f(a) is what you get when a is plugged into the equation, just as f(x) is what you get when x is plugged in to an equation.</p>

<p>So, let's say that in your question y = x + 2, and that the point (a,b) is at (2,3). Remember, the point is in the shaded region.</p>

<p>f(a) = 2+2 = 4.</p>

<p>here, statements 1 and 3 would be correct, and statement 2 wouldn't be.</p>

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<p>thanks dchow you are excellent math teacher...</p>

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<p>i feel sorry about my dumb explanations. i have even mistaken there to find 18 pi.</p>

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<p>How do you find the integral of 3x^3?</p>

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<p>A dance company plans to perform a piece that requires 2 dancers. If there are 7 dancers in the company, how many different pairs of dancers could perform the piece?</p>

<p>(I figured it out through the "dumb way"; counting out all the pairs, please tell me there is a faster way)</p>

<p>and sorry aisgzdavinci i haven't taken calculus yet, no idea</p>

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<p>I used combinations to get 21 (7!/(2!5!)). Is that right?</p>

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<p>I agree with you, trois.</p>

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<p>For aisgzdavinci, I guess it's 3x^4 / 4. If it's right then, I'll tell you how to do it.</p>

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<p>Another way of seeing that problem would be listing the dancers as ABCDEFG, then consider that there are 6 different combinations of dancer A (AB, AC, AD, AE, AF, and AG), 5 different combinations of Dancer B not including A (BC, BD, BE BF, BG), 4 different combinations of dancer C not including A or B (CD, CE, CF, and CG) and so on until you reach 1 combination of FG. Then sum up these combinations (6 + 5 + 4 + 3 + 2 + 1) to get 21.</p>

<p>I can't really explain it well.</p>

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<p>For ducktape14's Q i thought it was 42. Could sumone explain why it's 21. Im too thick to get the previous explanations :(</p>

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<p>Btw I have another Q, </p>

<p>Peter is assigning articles to his reporters. If he has 6 ideas and 4 reporters and each reporter can only write one article, How many different groups of articles can he assign?</p>

<p>Many thnx</p>

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<p>thank you for explaining but the answer is 18 pi....so what is the correction for 18pi
"p398 #11. I haven't done this one. But I guess I can do it.
It asks you to find the length of that curved path, not line AB.
If the circumference of the circle is 36pi, then radius will be R=18;
To find the length of that curved path, you should notice that if they complement each other, they make full circle, with radius of 18/4 = 4.5.
Then you should use formula 2 * pi * R to find the length, which is 9 pi."</p>

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<p>@ Humna. You should use the formula for this kind of problems. It's hard to write it here. Let me have an example for you.
Let's imagine you were to choose 2 friends from your 5 friends (A, B, C, D, E). Simply, you can do this tedious job, like AB, AC, AD, AE (Remember that it can be also written in the reversed order like BA, CA, DA, EA -- it doesn't make difference). But avoiding being perplexed, you should be careful.
AB, AC, AD, AE
BA, BC, BD, BE (omit BA -- it's included above as AB)
CA, CB, CD, CE (omit CA, CB)
DA, DB, DC, DE (omit DA, DB, DC)
EA, EB, EC, ED (all have been repeated)
So total choices would AB AC AD AE BC BD BE CD CE DE = 10 of them.
But you might not want to waste lots of time doing all these. So now it's time to use formula which has narrowed all these procedures.
n! / (k! * (n-k)!)
n is total something. k is number of something you should choose.
then for our example, it's 5! / 3! * 2! = 10;</p>

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<p>akati, have you even read my post? I told you how to get 18pi
Because line segment AB is the circle's diameter, it essentially divides the circle in half. Since they were asking for the length of the curved path from A to B, it makes sense that this curved path is half the whole circle. Since we already know the length of the whole circle, we just have to find the length of the half circle. How do we do that? By multiplying by 1/2, of course! So we get half of the circumference, which is 18pi</p>

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<p>Humna, the reason it's not 42 is due to the fact that all possible pairs will get counted twice if you listed every single pairing for the dancers. To account for this, we divide by half.</p>

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<p>@ Humna' 2nd problem:
You know one article can be written from one idea. So, if Peter has 6 ideas, then 6 different articles can be written. But there are only 4 reporters and each one can write only one article, which means there will be 2 ideas (or articles) left unused. But it's not important.
Let's imagine that Peter has A, B, C, D, E, F balls and has to put only one ball in 4 boxes in front of him. In the first box, he can put any of those 6 balls, so chances are 6. In the 2nd one, there are 5 left and chances are 5, and 4 chances for the 3rd box and 3 chances for the last (4th one). So, in total, 6 * 5 * 4 * 3 = 360.</p>

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<p>aisgzdavinci: The integral of 3x^3 is 3/4 x^4. This isn't even SAT math, it's calculus. What you do in these cases is you raise the exponent up 1 and then divide your answer by that number. So in 3x^3, the exponent is 3, add 1 to get 4, and then divide by 4. So your answer is 3/4 x^4. </p>

<p>Similarly, what is the integral of 2x^5? </p>

<p>It's 1/3 x^6, because you add one to 5, make that the new exponent, and then divide by 6. when you simplify you get 1/3 x^6. </p>

<p>When I tutor people I try to explain to them that there's always an explanation for these things. I could explain how this rule (called the power rule) works, but it would take time.</p>

<p>Humna's 2nd problem: "Peter is assigning articles to his reporters. If he has 6 ideas and 4 reporters and each reporter can only write one article, How many different groups of articles can he assign?"</p>

<p>Tsenguun, I think you misread the problem. "How many different groups of articles can he assign?" I take it to mean, "How many groups of 4 articles can be written from 6 ideas?" In this case, the answer would be 15, not 360.</p>

<p>For this we will use a simpler problem. We will use a problem that's less complicated so that you can see my idea.</p>

<p>If you had 3 shirts and 2 pairs of pants, how many possible outfits can you have?</p>

<p>6, because for each shirt you choose, there are 2 types of pants. 3 x 2 = 6.</p>

<p>Okay, now, back to the problem about the news articles. How many possibilities are there for the first article?</p>

<p>6.</p>

<p>How many possibilities are there for the 2nd article, after the first article has been chosen?</p>

<p>5.</p>

<p>How many are there for the 3rd article, after the first two have been chosen?</p>

<p>4.</p>

<p>How many for the 4th article, after the first three have been chosen?</p>

<p>3.</p>

<p>So, how many possibilities are there? 6 x 5 x 4 x 3, which is 360.</p>

<p>But the question asks, "How many different GROUPS can be assigned?" The group of articles A,B,C, and D is not different from the group of B,C,D, and A.</p>

<p>So we have to take that into account. How many ways can you arrange 4 articles? 4 x 3 x 2 = 12. Now you know that in your first answer (360), you've actually had (for example) ABC, and D written 12 different ways. And you've had B,C,D, and E written 12 different ways, and so on.</p>

<p>To get rid of these extra groups, we divide: 360/12 = 15. What happened was you had 15 groups, but they were each arranged 12 different ways. So we actually have 15 different groups.</p>

<p>Here's an excellent website: <a href="http://www.sparknotes.com/testprep/books/sat2/math1c/chapter11section3.rhtml%5B/url%5D"&gt;http://www.sparknotes.com/testprep/books/sat2/math1c/chapter11section3.rhtml&lt;/a> Please ask for help if you have any questions about this problem or any problem on the website.</p>

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<p>ouch!! hope i wouldn't make a careless mistake like this on my real SAT!!! misreading??? is it possible? kk</p>

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<p>It'sGr82BeAGator: Thank you...sorry for overglancing your response.</p>

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<p>Man I would be so totally lost without this website. Thnx to dchow--ur way better than my maths teacher at school. And tsengunn, man ur a maths genius. How much did u score on ur SAt? </p>

<p>Also thnk u, gr82beagator.</p>

<p>I dun mean to get sissy but kissses to all those who are smart and considerate enough to answer the Q. ;)</p>