<p>I got these questions wrong and don't know the correct ways to go about solving them: </p>
<p>1) Exactly 4 actors try out for the 4 parts in a play. If each actor can perform any one part and no one will perform more than one part, how many different assignments of actors are possible?</p>
<p>2) In the xy-coordinate plane, the graph x=(y^2)-4 interesects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?</p>
<p>1) This is how i do these questions: depending on how many different parts, i make 4 underscores, _ _ _ _ and fill each number up with 4, 3, 2, 1
4 actors can tryout for the first part, 3 actors try out for the second part (because one actor get the first part so he doesnt count), 2 actors tryout for the 3rd part (because another actor gets the 2nd part) ..etc get the drift?
so in the end you do 4X3X2X1= 24 assignments
is that the answer?</p>
<ol>
<li><p>24</p></li>
<li><p>1/5</p></li>
</ol>
<p>I would post explanations, but I don't know if they are correct</p>
<p>If you have a graphing calculator, then the problem seems a lot easier(for #2). Sadly, my graphing calculator is giving me problems, so I'll just give the algebraic way. Okay, so it said (0,5) and (5,t). So plug in 0 and 5 for x into the equation and you get two different y values(since the equation x=y^2-4 can be put into two equations because taking the square root of x+4 yields positive square root of x+4 and -negative square root of x+4). And then you should have 4 different points and should be able to find the slope from there.</p>
<p>same explanation for number 1 as ivyleague104</p>
<p>for number 2)
plug in 0 in place of x to find p.
x=(y^2)-4
0=(y^2)-4
4=(y^2)
y=p= 2</p>
<p>next plug in 5 in place of x to find t:
x=(y^2)-4
5=(y^2)-4
9=(y^2)
y=t=3</p>
<p>thus your two coordinates are (0,2) and (5,3)</p>
<p>find slope using slope formula
3-2/5-0 = 1/5</p>
<p>papakipari786, 4=y^2 can also be 2 AND -2, and for the 2nd one, can be 3 and -3. So what happens with -3 and 2?</p>
<p>The correct answer for the second one is 1.</p>
<p>(0, -2) and (5, 3)</p>
<p>3--2/5-0 = 5/5 = 1</p>
<p>I believe these are from the bluebook and I did remember something about using a the left negative one and the right positive one, which is what mvalenstein did. A useful tip to know is that if you ever have something like x=y^2, which you can't input into the calculator, you can make two separate square roots of X(negative and positive), and you can put those into the calculators and find the values of 0 and 5 on both lines.</p>
<p>sorry for the incorrect explanation.. i guess i made the same mistake as dudemanimcool</p>
<p>Problem 1 I understand now, I accidentally factored out the negative answer by being lazy and just algebraically changing the equation to y=(square root of: x + 4) on my calculator and going from there. I should have realized greatest implies multiple solutions... </p>
<p>Problem 2's answer was 18. I still don't know how you would solve it, the 4x3x2x1 way is how I went about it and apparently I got it wrong.</p>