Help with a math problem!

Test Preparation
1

<p>In the xy-plane, line l passes through the origin and is perpendicular to the line 4x + y = k, where k is a constant. If the two lines intersect at the point (t,t+1), what is the value of t?</p>

2

<p>Ok, so there is a lot going on in this problem, let’s break it down.</p>

<p>First rewrite the equation of the given line in y = mx + b form, this yields y = -4x + k</p>

<p>Using the information given in the problem about the other line, we can assume that the other line has slope 1/4 (since perpendicular lines have negative reciprocals for slopes) and intercept of 0 (since it passes through the origin).</p>

<p>We now have the equations y = -4x + k and y = (1/4)x</p>

<p>The next step is to solve for k. We know that both lines must have the point (t,t+1) so we can solve for k by plugging these values into the first equation. t + 1 = -4(t) + k. Therefore k = 5t + 1. </p>

<p>Now we can set these equations equal to each other. -4(t) + 5t + 1 = (1/4)t.
Solving this equation yields t = -4/3 and when we check our solution we get -1/3 from both equations.</p>

<p>(Note solving for k was an unnecessary step but I thought it might help you see how the problem works. Hope this helps.)</p>

3

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<a href=“http://i40.■■■■■■■.com/10zmu5h.jpg[/IMG]”>http://i40.■■■■■■■.com/10zmu5h.jpg

</a></p>

<p>Which of the following graphs best represents the information in the table above? Why?</p>

4

<p>Chart A describes the data best. </p>

<p>If you look at the data in the table above you can see that initially there is a large amount of growth between weeks 4 and 5 and 5 and 6 that then slows between weeks 6 and 7 and 7 and 8. Looking at the charts this trend best described by A, fast growth initially and slow growth thereafter. B shows slow initial growth that is followed by fast growth later on.</p>

5

<p>When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?
Options: 1,2,3,4,5</p>

6

<p>1, 2, and 3 don’t work (since remainders are defined to be strictly less than the divisor itself).</p>

<p>Clearly 15 is divisible by 5. Also 15/4 gives a remainder of 3. So only 4 works → answer is 1.</p>

<p>Edit: Sorry misinterpreted the answer choices as being the only options for k. 15/6 and 15/12 work. Thanks Sikorsky.</p>

7

<p>The way I read this question, 1, 2, 3, 4 and 5 aren’t options for the divisors. They’re options for how many values of k.</p>

<p>15/1 has remainder 0. So have 15/3, 15/5 and 15/15.</p>

<p>15/2 has remainder 1. So has 15/7.</p>

<p>You get a remainder of 3 when you divide 15 by 4 or 6 or 12.</p>

<p>There are 3 possible values for k.</p>

<p>@MITer–I don’t think that was worded very clearly. I base my interpretation more on the question I would ask (and the questions I’ve been asked in the past) than on the words on my screen.</p>

8

<p>@ 110percentwahoo wow you’ve explained it so clearly! thanks a lot!</p>