<p>Three lines are drawn in a plane so that there are
exactly three different intersection points. Into how
many nonoverlapping regions do these lines divide the
plane?
(A) Three
(B) Four
(C) Five
(D) Six
(E) Seven </p>
<ol>
<li> A club is buying boxes of candy bars to sell for a fundraiser. If each box contains c candy bars, and each
member sells x bars each day, how many boxes are
needed to supply enough candy bars for 3c members
to sell for 5 days?
(A) 15c^2x
(B) x/15
(C) 3x/5
(D) 15c^2/x
(E) 15x</li>
</ol>
<p>I think the answer to each question is E.</p>
<p>Three lines are drawn in a plane so that there are
exactly three different intersection points. Into how
many nonoverlapping regions do these lines divide the
plane?
(A) Three
(B) Four
(C) Five
(D) Six
(E) Seven </p>
<p>For this one draw three lines that make a triangle in the middle; that way, the three lines will intersect at three points. Then, count the number of regions that this triangular thing makes. It would be seven.</p>
<ol>
<li>A club is buying boxes of candy bars to sell for a fundraiser. If each box contains c candy bars, and each
member sells x bars each day, how many boxes are
needed to supply enough candy bars for 3c members
to sell for 5 days?
(A) 15c^2x
(B) x/15
(C) 3x/5
(D) 15c^2/x
(E) 15x </li>
</ol>
<p>For this one, try to ignore the “3c” member part. Just think that it is “n” number of members. So “n” number of members will sell “xn” bars each day; for 5 days, they will sell total “5xn” bars. Now that you won’t be confused with “3c,” plug it in instead of “n.”
You will get “15xc.” Finally, divide this formula by “c” to find the total number of boxes you will need. The answer is 15x, E. </p>
<p>Hope this helped you :)</p>
<p>^Ahh that helped a lot, thanks Jenny!</p>
<ol>
<li>A club is buying boxes of candy bars to sell for a fundraiser. If each box contains c candy bars, and each
member sells x bars each day, how many boxes are
needed to supply enough candy bars for 3c members
to sell for 5 days?
(A) 15c^2x
(B) x/15
(C) 3x/5
(D) 15c^2/x
(E) 15x </li>
</ol>
<p>I would pick numbers for this one. For example, if we let c=10 and x=2. Then there are 3<em>10=30 members. We then need 30</em>5*2=300 candy bars, or 300/10=30 boxes.</p>
<p>Now substitute our values of c and x into each answer choice:</p>
<p>(A) 15c^2x=15(10)^4
(B) x/15=2/15
(C) 3x/5=6/5
(D) 15c^2/x=15(10)^(2/2)=150
(E) 15x=15*2=30</p>
<p>Since only choice (E) came out correct, that is the answer.</p>
<p>that number 19 looks confusing!! is this a question that has been on the act or sat?</p>
<p>^Yeah its from an online course praactice test I believe.
And thanks Dr.Steve!</p>
<p>I believe 19 was on an actual SAT as the last question of a math section.</p>