Help with some Math problems!

Test Preparation SAT Preparation
1

<p>You know you love doing this.</p>

<ol>
<li>If the sum of the consecutive integers from -22 to x is 72, what is x?
A. 23
B. 25
C. 50 (Answer)
D. 75
E. 94</li>
</ol>

<p>I thought it was 94 (E). Can anyone explain why it's C?</p>

<ol>
<li><a href="http://i.imgur.com/EGJAS.jpg%5B/url%5D"&gt;http://i.imgur.com/EGJAS.jpg&lt;/a&gt;&lt;/li>
</ol>

<p>Answer is A. No idea how to do this. I assume you plug in numbers?</p>

<ol>
<li><a href="http://i.imgur.com/DqYHz.jpg%5B/url%5D"&gt;http://i.imgur.com/DqYHz.jpg&lt;/a>
Answer is E.</li>
</ol>

<p>I'll be back in a few with more!</p>

2

<p>The answer to 18 is B (25).</p>

<p>-22 to 22 cancels out. 23+24+25 = 72.</p>

3

<p>20.</p>

<p>You have to know how to shift functions.</p>

<p>h=-3 (shift right three)
k=-2 (shift down two)</p>

<p>-3 x -2 = 6</p>

4

<p>@LoseYourself: Well, now I feel dumb. Thanks.</p>

<p>@LoseYourself(again!): Thanks. I’m going to look that concept up too.</p>

5

<ol>
<li>You can plug in numbers for k and n, but you will still have to do some algebra. So here is an algebraic solution:</li>
</ol>

<p>x^(-2/3) = k^(-1) = 1/k (I raised both sides of the first equation to the 1/2 power)</p>

<p>y^(-2/3) = n^(-1) = 1/n (I raised both sides of the second equation to the -1/2 power)</p>

<p>(xy)^(-2/3) = x^(-2/3)y^(-2/3) = (1/k)(1/n) = 1/(kn) = 1/(nk)</p>

<p>Remark: Notice that you “undo” a power by raising to the reciprocal power.</p>

6

<p>@DrSteve: Awesome. I don’t know how you guys make it seem so easy. Thanks!</p>

7

<p>One more!</p>

<p><a href=“http://i.imgur.com/kbV0B.jpg[/url]”>http://i.imgur.com/kbV0B.jpg&lt;/a&gt;
Answer is D. Why?</p>

8

<p>One more!</p>

<p><a href=“http://i.imgur.com/kbV0B.jpg[/url]”>http://i.imgur.com/kbV0B.jpg&lt;/a&gt;
Answer is D. Why?</p>

9

<p>Label the vertex of the cube beneath B as point C.</p>

<p>Label the vertex of the cube to the right of A as point D.</p>

<p>What you are looking for is the hypotenuse of the right triangle ABC. In order to find it you first have to determine the hypotenuse of triangle ACD.</p>

<p>You know that AD is 1, and DC is 2. So the hypotenuse of ACD (line AC) is the sqr rt of 5 (AD^2 + DC^2 = AC^2).</p>

<p>Now you have BC = 1 and AC = sqr rt of 5. So the hypotenuse of triangle ABC (line AB) is the sqr rt of 6 (BC^2 + AC^2 = AB^2).</p>

10

<p>I tried figuring it out on my own. I already took SAT twice and there seemed to be no questions or maybe only 1 or 2 this hard.</p>

<p>I eventually worked out the right answer, but only since I knew the answer in the first place.</p>

<p>anymore? keeps me sharp for the upcoming fall testing!!!</p>

11

<p>Haha. That’s really not that hard.</p>

<p>Simply put: you need to find both legs.</p>

<p>1st leg: you know it’s 1 because it’s the midpoint of a side.</p>

<p>2nd leg: you have to solve a triangle with legs 1 and 2. equals (5)^(1/2)</p>

<p>You got your legs. Plug into Pythagorean formula to get answer. :D</p>

12

<p>Awesome! I don’t have any more Math questions for now (tomorrow’s CR time), but when I do I’ll be back! Hopefully I’ll have less questions next time.</p>

<p>You guys are great help!</p>

13

<p>Okay one more. The only problem is that I don’t have answer for this one! :frowning: Wasn’t on the answer key.</p>

<p><a href=“http://i.imgur.com/DwDIL.jpg[/url]”>http://i.imgur.com/DwDIL.jpg&lt;/a&gt;&lt;/p&gt;

14

<p>x+2/2 is the average of x+y. This means that x and y each have to be four units away from the average. This makes the answer E.</p>

15

<p>For the first one with the cube, just use the Generalized Pythagorean Theorem (this finds the “long diagonal” of a rectangular solid):</p>

<p>d^2 = a^2 + b^2 + c^2 = 1^2 + 2^2 + 1^2 = 1+4+1=6
So d is the square root of 6, choice (D).</p>