<p>Checked the consolidated list of solutions already, and I need help with a few problems:</p>
<p>P409 #7.
Dwayne has a newspaper route for which he collects k dollards each day. From this amount he pays out k/3 dollars per day for the cost of the papers, and he aves the rest of the money. In terms of k, how many days will it take Dwayne to save $1,000?</p>
<p>P412 #15.
This one has a figure, so anyone with a blue book please help!</p>
<p>P398 #13.
3, 5, -5, ...
The first term in the sequence of numbers shown above is 3. Each even-numbered term is 2 more than the previous term and each odd-numbered term, after the first, is -1 times the previous term. What is the 55th term of the sequence?</p>
<p>p. 409, #7
Savings = x [days] * k [$/day] - x [days] * k/3 [$/day] = 1000 [$]
xk - (xk)/3 = 1000
[3xk - xk]/3 = 1000
2xk = 3000
x = 1500/k</p>
<p>p. 412, #15
If I can find that page from the blue book (some of my pages are torn out), I'll try to help w/ that one.</p>
<p>p. 398, #13
There is a pattern here. Note that if you pick any nth term and add 4 terms (in other words, say you picked the first term. When I add 4 'terms,' I mean to say 1 (first) + 4 = 5 (fifth term). Sorry for lack of better wording.), you end up with the same value!</p>
<p>n1 = 3 <--- Note that this term...
n2 = 5
n3 = -5
n4 = -3
n5 = 3 <--- ...has the same value as this term!
n6 = 5
n7 = -5
n8 = -3
n9 = 3 <--- Same!</p>
<p>So, to get started, you might write an equation to represent each time you'll get a value of 3. To do this, I would use f(x) = 1 + 4x, where x is an integer. To test it, plug in easy integers for x. f(1) = 1 + 4(1) = 5. f(2) = 1 + 4(2) = 9 and so forth.</p>
<p>If you plug in 13 (by guess and test) for x in that function, you get f(13) = 1 + 4(13) = 53, or the 53rd term. We want the 55th term, so just proceed down the pattern two more terms, and you'll get the 55th term's value, which should be -5.</p>
<p>I am sure that is more compact and efficient way to do this type of problem, but I have never attempted to take that path in this scenario yet, lol.</p>
<p>Have you been taught modular / "clock" arithmetic yet? It goes something like this...
x/y can be rewritten w/ mod to express the remainder of the division of x by y. x mod y = remainder</p>
<p>The remainder can be useful to find the original / first term that has the value you desire. x = term asked of, and y = # of terms that have to be passed over until you arrive at the same value again.</p>
<p>So, for this scenario, we have x mod 4 = simplest/smallest answer possible
where x will now = 55 and
55 mod 4 = 3 (b/c 55/4 yields a quotient of 52 and a remainder of 3. The quotient doesn't mean anything for us w/ this concept.)</p>
<p>This result tells us to look at the third term, and so we will! n3 = -5, which is also what n55 is equal to.</p>
<p>okay, this is going to sound so stupid..but for #16 on page 412..im just not good with math logic, can anyone help me out? i know it's easy, but i just cant figure it out!..blue book by the way</p>
<p>all you have to do is substitute into the equations. If you tend to get confused between variables and what not, the best way to stay organized is to put parantheses around everything as you substitute. for example:</p>
<p>4 triangle 5y = 5y square 5</p>
<p>substituting definitions of triangle and square</p>
<p>For those types of problems, if you're not too good on figuring things out with variables, try substituting numbers.</p>
<p>Say that he collects $90 (k value) a day. That means he pays out $30 (90/3) for the cost of papers. So that means he makes $60 profit each day. Divide 1000/60 = 16.67 approximately. Plug in the K value and see which one gives you 16.67.</p>
<p>This method may not be so time saving in this situation, but for even harder problems, plugging in your own values helps out a lot =)</p>