  # SAT Help(And Freakout)!!!!

<p>Hey, so I have these two questions that I've been trying to figure out all night. Any help is appreciated.</p>

<ol>
<li>The cube shown above has edges of length 2, and a and b are midpoints of two of the edges. What is the length of line AB.</li>
</ol>

<p>a. Square root of 2
b. Square root of 3
c. Square root of 5
d. Square root of 6
e. Square root of 10</p>

<ol>
<li>Let x be defined as x= x^2-x for all values of x. If a= a-2, what is the value of a?</li>
</ol>

<p>a. 1
b. 1/2
c. 3/2
d. 6/5
e. 3</p>

<p>We need the figure of 15 to solve it. And 16 makes no sense.</p>

<p>For 15, I cannot know the answer unless I have the picture or you have to describe where A and B are located.</p>

<p>For 16, I believe the answer is C. Plug a into the equation and you get a = a^2 - a. Then plug a-2 into the equation and you should get a^2 - 5a + 6. Set that equal to a^2 - a and a should equal 3/2.</p>

<p>@ Entropyrising: Even though a = a-2 is nonsensical, it makes sense once you plug it into x = x^2 - x.</p>

<p>a=a-2
(3/2)=(3/2)-2
3/2=-1/2
False</p>

<p>It doesn't make sense no matter what.</p>

<p>Also (a-2)^2-a is a^2-5a+4</p>

<p>If a^2-5a+4=a^2-a
-5a+4=-a
4=4a
a=1</p>

<p>Also, if a^2-5a+4=a^2-a and a=a-2, then
(a-2)^2 -5(a-2)+4=a^2-a
a^2-4a+4-5a+10+4=a^2-a
a^2-9a+18=a^2-a
-9a+18=-a
18=8a
9/4=a</p>

<p>Makes no sense</p>

<p>Well A is located in the middle of the bottom face of a 3D square. B is located in the middle of the right face of the 3D square.</p>

<p>So it is basically a right triangle with legs 1 and 1.
Use Pythagorean Theorem.
2^.5</p>

<p>But the answer is Squae root of 6 :(</p>

<p>@Entropyrising: a=a-2 is does not make sense, but plugging into the x equation does. Obviously this is a terrible (and will not appear on the actual SAT) question but it's testing algebraic manipulation and number theory skills.</p>

<p>@OP: Most like you described the question wrong. Are the points located on edges or faces?</p>

<ol>
<li><p>Actually the question was asked in the Official SAT Book, so it could be asked. Anything goes.</p></li>
<li><p>The points are midway on the edges. Bottom and Right</p></li>
</ol>

<p>a=a-2 does not make sense and plugging it into the equation is a false manipulation. And you can plug it in anywhere you want. You had no specific reasons to plug them in where you did.</p>

<ol>
<li>If I remember correctly the answer is root 6 and if memory serves further this question is in the math section of test 7 of the BB 2nd Ed.</li>
</ol>

<p>To solve, you do the Pythagorean theorem twice. The first to solve for the hypotenuse of length from the top corner of point B to the base of A. 2^2 + 1^1 = 5. So the hypotenuse of the the first right triangle is root 5. Root 5 is now one of the legs of the new triangle. Plug it in again. 5 (the square and root cancel) + 1^1 = 6. Take the root of that so root 6. </p>

<ol>
<li>I think this one is the one with the with functions in ovals. (So it's classified as a funny symbols equation)
Just plug in the (a-2) for all values of "x" so x= x^2-x becomes</li>
</ol>

<p>(a-2) = (a-2)^2 - (a-2).</p>

<p>It works out to be 3/2.</p>

<p>16 makes much more sense if it reads: f(x)=x^2-x and then f(a)=f(a-2)</p>

<p>I don't have the blue book in front of me but that's how I remember it...besides, as has been pointed out, "let a=a-2" makes no sense mahematically (though it is a legal instruction in some computer languages).</p>

<p>Haha yea it is f(x)=x^2-x and then f(a)=f(a-2)</p>

<p>:P</p>

<p>Never mind, I got the answer ;)</p>