<p>After an extensive explaination of Xiggi’s 2(sp1*sp2)/(sp1+sp2) I have to ask, are there any more special formulas that can help me on the SAT? I learned one more on my own.</p>
<p>You know those consecutive questions? Say “3 odd con. #s at up to 111. What is the smallest number?”</p>
<p>111/3= 37<– so this is the one in the middle
35+37+39=111</p>
<p>The answer is 35. That little math trick helped so much!</p>
<p>Or you could do it:</p>
<p>n = smallest number.</p>
<p>=> n + (n + 2) + ((n + 2) + 2) = 111</p>
<p>=> 3n = 105</p>
<p>=> n = 35</p>
<p>:)</p>
<p>Either way works.</p>
<p>HiWei, do you have anymore? NOt just for this particular type of problem but for any others on the SAT.</p>
<p>Um… I havent taken the SAT in a while… the tricks usually come to me during the test… If you have a specific problem, I could probably help.</p>
<p>:)</p>
<p>In the xycoordinate system (square of 6, k) is one of the points of interesction of the graph y=x^2-7 and y = -x^2+j?
a. 5
b. 4
c. 3
d. 2
e. 1</p>
<p>Another question.
The first 3 terms of a sequence are 3,9,27… and each additional term is found by multiplying the preceding term by 3.</p>
<p>Answer:
-look for the patter by finding a few more terms.
-look at the patter of the units digit. If you find the pattern, the units digit pattern is 3,9,7,1,3,9,7,1
<p>Now, that 3rd part makes no sense. How do you know that the 48th number will have the 4th number?</p>
<p>because 48 is evenly divisible by 4</p>
<p>Its also evenly divisible by 2 and 1, but 9 or 3 is not the answer.</p>
<p>The units digits in the series repeat in the pattern 3,9,7,1,3,9,7,1 and so on. This series repeats every 4 terms, so to find the units digit of the nth term, divide n by 4 and find the remainder. If the remainder is one, the units digit is 3. If r = 2, units = 9. If r = 3, units = 7. If r = 4 = 0, units = 1.</p>
<p>Rs are
0,2, and 3? What happened to 1?</p>
<p>here’s another SAT math trick: on a hard problem, try to do it the “non-math” way. there always is an algebraic solution to a problem, but you might be able to get it much quicker and more easily by skipping the math and just using common sense.</p>
<p>bumpityhump</p>
<p>Sorry… I wrote out “remainder” the first time, and then shortened it to “r”.</p>
<p>No I am talking about the remainder. You didn’t say what happens if the reminader is 1.</p>
<p>“If the remainder is one, the units digit is 3” such as in the 1st, 5th, 9th etc terms</p>
<p>Oh hah sorry.</p>
<p>
</p>
<p>Okay, I am going to assume that “square of 6” means the “square root of six.” Remember, there is a big difference. I am also going to assume the question asks for the value of j. Now since you know that this point (sqrt6, k) is the intersection of the two graphs, you should be able to substitute sqrt6 in for x in both equations, and k in for y in both equations. Substitute into the first equation and you solve to find out that y=-1, and thus, k=-1.</p>
<p>Next, substitute -1 for y into the next equation. Put in sqrt(6) for x. Solve for j. Remember to watch the order of operations between the squared, and the negative in this final step.</p>
<p>I’m curious to see what exactly that question is asking myself. If you know that the two equations have the same x and y coordinates at a particular point, you can just set the equations equal to each other and plug in the x-value (square root of 6) to solve for j. In this case, j = 5, which is also what you’d get by solving for y first, then plugging in to the 1 equation, etc.</p>
<p>bumpityhumphump</p>
<p>I have also began to notice that questions that look hard can actually be very simple. Like the one I posted. Infact, whenever you brainiacs answer a question, there is a way to get it that I can understand (after a little digging)</p>
<p>I want to know, do any of you have a way that you deal with questions with equations that are mostly letters. Where the answer is “What is blah in terms of k and a?”</p>
<p>What I have always done was plug in answer choices, or assign a number for the variables.</p>
<p>Plugging in does work very well a lot of the time. However, a quicker method would need a specific question.</p>