SAT math from blue book assistance pLEASE!

Test Preparation SAT Preparation
1

<p>This is from test five from the blue book.</p>

<p>Section 3 </p>

<ol>
<li>The Acme Plumbing Company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consist of 1 experienced plumber and 2 trainees, how many different such teams are possible?</li>
</ol>

<ul>
<li>I thought this was straighforward. 4<em>4</em>3=48. but the book says im wrong...why?</li>
</ul>

<p>Same section </p>

<ol>
<li></li>
</ol>

<p>h(t)=c-(d-4t)^2
At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at the time t=2.5, what was the height, in feet, of the ball at time t=1.</p>

<p>*how the hell do i solve this?</p>

<p>Section 7</p>

<ol>
<li>If k,n,x and y are positive numbers satisfying x^(-4/3)=k^(-2) and y^(4/3)=n^2, what is (xy)^(-2/3) in terms of n and k ?</li>
</ol>

<p>I thought the answer was (c) k/n</p>

<ol>
<li>This one has a figure and it asks about f(x) and g(x) and such. How would u solve such question?</li>
</ol>

<p>THANK YOU</p>

2

<ol>
<li>The way you’re doing it assumes that a team of Trainee 1 and Trainee 2 is different from a team of Trainee 2 and Trainee 1 (when in fact both are the same). Since this problem isn’t too complicated, write out the possibilities.</li>
</ol>

<p>Let experienced plumbers be A, B, C, and D
Let trainees be 1, 2, 3, 4</p>

<p>4 possibilities for experienced plumbers (duh)</p>

<p>possibilities for a two-man trainee team:
1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4
which is a total of 6</p>

<p>4 x 6 = 24</p>

<ol>
<li>This is just algebraic manipulation
h(t) = c - (d - 4t)^2</li>
</ol>

<p>At time t=0, a ball was thrown upward from an initial height of 6 feet.
h(0) = 6 = c - (d- 4*0)^2
6 = c - d^2</p>

<p>If the ball reached its maximum height of 106 feet at the time t=2.5
h(2.5) = 106 = c - (d - 4*2.5)^2
106 = c - (d - 10)^2</p>

<p>what was the height, in feet, of the ball at time t=1.
106 = c - (d - 10)^2
106 = c - (d^2 - 20d +100)
106 = c - d^2 + 20d - 100 but 6 = c - d^2 so
106 = 6 + 20d - 100
20 d = 200
d = 10</p>

<p>If you look at the equation, you realize that this height model is just an upside down parabola. If you know your equations of parabolas, you know that c is the maximum value (the vertex, 106). So c = 106. You could’ve used this fact to solve for d much more easily, but I wrote everything out for clarity.</p>

<p>h(t) = c - (d - 4t)^2 becomes
h(t) = 106 - (10 - 4t)^2
h(1) = 106 - (10 - 4)^2
h(1) = 70</p>

<ol>
<li>If k,n,x and y are positive numbers satisfying x^(-4/3)=k^(-2) and y^(4/3)=n^2, what is (xy)^(-2/3) in terms of n and k ?</li>
</ol>

<p>(xy)^(-2/3) = x^(-2/3) y^(-2/3)</p>

<p>How do we manipulate x^(-4/3)=k^(-2) to get x^(-2/3)? Raise each side to the 1/2 power:
[x^(-4/3)]^(1/2) = [k^(-2)]^(1/2)
x^(-2/3) = k^(-1)</p>

<p>How do we manipulate y^(4/3)=n^2 to get y^(-2/3)? Raise each side to the -1/2 power:
[y^(4/3)]^(-1/2) = [n^2]^(-1/2)
y^(-2/3) = n^(-1)</p>

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

3

<p>very informative, i think you should be paid to do this. thanks</p>

4

<p>Im still somewhat confused about the first one</p>

<p>You have four possible experienced so its 4. now for the second spot there are four available trainees and for the last spot there would be only 3 available trainees.
so 4<em>4</em>3…?? I see that 24 is right, but this is how I solve most of problems like this</p>