<p>i had problems solving these questions (i either got em wrong or right but just not sure)</p>
<p>can anyone please explain how to solve the questions (briefly, as many as possible)</p>
<p>Problems from REAL SAT II (collegeboard)</p>
<li><p>If X0= 0 and Xn+1 = square root(6+Xn), then X3=
A) 2.449
B) 2.907
C) 2.984
D) 2.997
E) 3.162</p></li>
<li><p>The radius of the base of a right circular cone is 6 and the radius of a parallel cross section is 4. If the distance between the base and the cross section is 8, what is the height of the cone?
A) 11
B) 13.333
C) 16
D) 20
E) 24</p></li>
<li><p>Suppose the graph of f(x)=-x^2 is translated 3 units left and 1 unit up. If the resulting graph represents g(x), what is the value of g(-1.6)?
A) 2.96
B) -0.96
C) -1.56
D) -1.96
E) -2.56</p></li>
<li><p>In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?
A) 120
B) 210
C) 240
D) 5040
E) 14400</p></li>
<li><p>What is the length of the major axis of the ellipse whose equation is 60x^2 + 30y^2 = 150?
A) 1.26
B) 2.5
C) 3.16
D) 4.47
E) 5</p></li>
</ol>
<p>If X0=0 and Xn+1=sqrt(6+Xn), then X1=X0+1=0+sqrt(6+0)=sqrt(6)
It then follows that X2=X1+1=sqrt(6+sqrt(6)).
X3 is then X2+1=sqrt(6+(sqrt(6+sqrt6))=2.984.</p>
<p>I'm on to #2 now. Hopefully the above was clear.</p>
<p>I'm not so sure about this one (as I haven't done geometry in a while), but the heights are in proportion to the radii. Thus, the height of the top of the cone to the parallel cross-section is 2/3 of 8, or 5.3333. Add to 8 and you have 13.333. Again, I'm not sure if you can do this.</p>
<p>Suppose the graph of f(x)=-x^2 is translated 3 units left and 1 unit up. If the resulting graph represents g(x), what is the value of g(-1.6)?
f(x) = a(x-p)^2 + q
The inverse of p, and the value for q make up the x and y coordinates of the vertex. Since the graph is translated 3 units left and 1 unit up, that would make the equation
f(x) = (x+3)^2 + 1
That makes the vertex (-3, 1).</p>
<p>Substitute (-1.6) into the equation to get g(-1.6)</p>
<ol>
<li><p>If X0= 0 and Xn+1 = square root(6+Xn), then X3=
X3 = sqrt(6+X2) = sqrt(6+sqrt(6+X1)) = sqrt(6+sqrt(6+sqrt(6+X0)))
Plug in X0=0 for the answer.</p></li>
<li><p>The cross-section radius decreases linearly as you go upwards from the base. Formally,
cross-section radius = base<em>radius * (cone</em>height - distance<em>above</em>base)/ cone<em>height
or cross-section radius = base</em>radius * (1 - (distance<em>above</em>base/ cone_height))
4 = 6 (1 - 8/h)
4/6 = 1-8/h
8/h = 1-(2/3) = 1/3
h = 24</p></li>
</ol>
<p>(A quicker way is to notice that if you went up 8 units from the base, the radius dropped by 2, from 6 to 4. Go up another 8 units, and the radius will drop by another 2, to 4-2=2. Go up yet another 8 units, and the radius will be 2-2 = 0; you're at the top of the cone. Total distance travelled from base = 8+8+8 = 24).</p>
<ol>
<li><p>Do it in stages. If the graph 'moves' 3 units to the left, the new function (say h(x)) will be h(x) = -(x+3)^2. If it now 'moves' up 1 unit, the function g(x) will be 1 + h(x) = 1 - (x+3)^2.
Plug in x= -1.6, and compute g(-1.6) = 1 - (3 -1.6)^2 .</p></li>
<li><p>You just need to find the #different ways of getting a 3 person group, the remaining people automatically form a 7 person group. The answer is nC3 = 10! / (7! 3!) = (10)(9)(8) / ((3)(2)) = 120.</p></li>
<li><p>At x=0, y^2 =150/30 or y = +/- sqrt(5)
At y=0, x^2 =150/60 or x = +/- sqrt(2.5)</p>
<p>y is the major axis, and its length is from -sqrt(5) to +sqrt(5)
= 2 sqrt(5), which should be 4.47</p></li>
</ol>
<p>optimizerad ure juz amazing thank u so much
thank you so much everyone!!</p>
<p>juz one more question!!
Which of the following has an element that is less than any other element in that set?
I. The set of positive rational numbers
II. The set of positive rational numbers r such that r^2 >= 2
III. The set of positive rational numbers r such that r^2>4</p>
<p>The answer is None; can anyone explain this??</p>
<p>The least set of positive rational numbers doesn't have a least element. Assume it does, call it a/b. Now consider a/2b, which is also rational. Contradiction, so there is no least element. You can find similar contradictions for the other ones.</p>
<p>for your latter question, set I is out because rational numbers can go to infinity and still be positive (1/1000, 1/100000, etc). Set II and III are the same; (201/100)^2 is bigger than 4 as well as (20000001/10000000)^2, etc.</p>