<p>If x and y are numbers such that (x+9)(y-9)=0 what is the smallest possible value of x^2+y^2 ?</p>
<p>a.)0
b.)9
c.)18
d.)81
e.)162</p>
<p>Alright I may not be right since I’m not very good at math. But this is my thinking. Either X is -9 or Y is 9 in order to satisfy the equation. So, if Y was to be 9, then Y squared would be 81. You keep X at 0 so zero squared is 0. 81+0=81.</p>
<p>Correct me if I’m wrong.</p>
<p>another one…</p>
<p>The original price of a book was* b* dollars. This price was discounted x percent during a sale. James used a coupon to buy the book fort percent off the sale price. Which of the following represents the price, in dollars, that James paid for the book?
(A) b[1-(x/100)]<a href=“B”>1-(t/100)</a> b[(1-x)(1-t)]/100
(C) b<a href=“D”>1-((x+t)/100)</a> b<a href=“E”>1-(xt /100)</a> b[(x/100)(t/100)</p>
<p>To solve this problem, I did this
b-(x/100) this is the part when x% was discounted from the original price.
Then, I did [b-(x/100)]-[(t/100)(b-(x/100))] to show the amount of money subtracted from the sales price… I think this is where I went wrong, but I don’t know why. Can someone please explain this?</p>
<p>I’m not sure but I think the answer is C. If you plug (x,y) in as either (-9,0) or (0,9)…</p>
<p>kobudnik, you’re not bad at math. lol you’re right. Could you explain a bit more?</p>
<p>
</p>
<p>It’s probably a silly mistake. 
You should have written [b-(**b**x/100)] instead of just [b-(x/100)]. You’re can’t just subtract x% from b, you have to subtract x% of b from b.</p>
<p>When I first looked at the problem I realized one of the variables canceled out one of the constants to equal 0. Either x would be -9 or y would be 9.</p>
<p>
</p>
<p>Eqn: (x+9)(y-9)=0
Divide by: (y-9)
Now: (x+9)=0
x = -9</p>
<p>or</p>
<p>Eqn: (x+9)(y-9)=0
Divide by: (x+9)
Now: (y-9)=0
y = 9</p>
<p>You can use y=9 or x=-9. Anyways from here you should look back at the problem at it says you need to plug in x and y into the equation x^2 + y^2 = lowest number possible. Now you realize y either has to equal 9 or x has to equal -9 while the other variable can equal zero. (0)^2 + (9)^2 = 81</p>
<p>Now if you didn’t know how to completely solve it you can cancel answers. You know x and y both arent equal to zero. (a is gone). You know either a or b isn’t equal to 3 (b and c are gone). With simple reasoning you can easily cross out answers a, b, and c.</p>
<p>Collegeboard is trying to trick people w/ answer E since they want people to plug in x=-9 and y=9.</p>
<p>Correct answer = D.</p>
<p>I suggest any equation that involves money or percentages to make up numbers for variables because you can simply check your work.</p>
<p>
</p>
<p>Let b = $1000
Let x = 10%
Let t = 25%</p>
<p>$1000(1-.1) = $1000(.9) = $900
$900(1-.25) = $900(.75) = $675</p>
<p>Just by looking at the problem I can see (A) looks very similar to my setup.</p>
<p>Plugin (A):
=1000[1-(10/100)][(1-(25/100)
=1000[1-(.1)][(1-(.25)]
=1000(.9)(.75)
=1000(.675)
=675</p>
<p>Nevermind I’m stupid</p>
<p>Everyone, thank you so much for helping. Kobudnik, Crouch88, and Avidstudent, thank you. :)</p>