Math Questions

Test Preparation SAT Preparation
1

<p>Here I gathered some math questions that I do not know how to solve from the officaial study guide (BB),2nd edition.</p>

<ol>
<li><p>page 401-19
It's a pyramid that has altitude h and a sqaure base of side m. The 4 edges of the square meet at the vertex of the pyramid, V. The length between the 4 edges to vertex V is e. And e=m. So what is the value of h (altitude) in terms of m?
(There is a figure onpg 401 if the info above is not clear enough)</p></li>
<li><p>page 548-16
A cube with volume 8 cubic centimeters is inscribed in a sphere so that each vertex of the cubic touches the sphere. What is the length of the diameter, in cm, of the sphere?
(the answer it was given is not what I worked out,and I doubt if the answer is right.)</p></li>
<li><p>page 598-18
There are 5 cards placed in a row, and one of them is not at either end, so how many different arrangements are possible?</p></li>
</ol>

<p>I would really appreciate anyone could help me out here. Also, just a general question, how can I stop making stupic mistakes in the math section?</p>

2

<ol>
<li><p>Draw a line from the right vertex (or any vertex) of the pyramid to the bottom of ‘h’
You will see that you now have a right triangle with m as the hypotenuse, and h and
mROOT(2)/2 as the legs. Applying Pythagorean Theorem (solve for h) to the triangle
will yield m/ROOT(2) as the correct answer.</p></li>
<li><p>Cube’s edges will each be 2. Since the diameter has to run thru the center, you must
Use the legs of (2 and 2Root(2)) and solve for the hypotenuse. Which simplifies to
2 Root(3)</p></li>
</ol>

3

<ol>
<li>Total of 5! choices. Subtract the choices when card is at either end 2(4!) and you will
be left with 72.</li>
</ol>