<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>
<p>A) -4/3
B) -5/4
C) 3/4
D) 5/4
E) 4/3</p>
<p>When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?</p>
<p>A) One
B) Two
C) Three
D) Four
E) Five</p>
<p>Also if anyone has the blue book #16 out of practice Exam 3 Section 8</p>
<p>any help greatly appreciated</p>
<p>Georgia Tech? For the first one change the first equation form standard to slope-intercept form. From their find the slope of that line and take its negative reciprocal. Then you set the y=mx+b (b is 0 since it goes through origin). Then you just put in the k values into the equation. The second one might be easiest for you to just go through trying all variables between 3 and 13 exclusive.</p>
<p>i think u got the test and section wrong, but anyways lets look at 2 first.</p>
<p>C) 3 should be the answer.</p>
<p>Long way:
Think about this logically. if its positive, must be above 0, and u can cancel out 1, 3, 5, 15. Because they are multiples of 15. If you think about the other numbers…12 works..goes into 15 1 time R3. 14, 13, 11, 10, 9, 8, 7 dont work. 6 works, 5 doesnt 4 works. and thats it…</p>
<p>4, 6 and 12 work. thats 3…</p>
<p>Short way:</p>
<p>If the remainder is 3, the “number” has to be 3 away from 15, and considering its division, it cant be 3 above 15…so it has to be 12. Now just concider the factors of 12… 1-12, 2-6, 3-4. Immediatly u can eliminate, 1, 3 because they go directly into 15. all u have left to eliminate is 2 because 2*7=14, remainder is 1, and ur left with 4, 6, 12</p>
<p>Ok for the 1st one, i think the answer is A. This is how i did it.</p>
<p>if 4x+y=K the y=-4x+k; which means line L = 1/4x=y (neg. recip.)</p>
<p>If the slope is 1/4, and we know that line L passes through (0,0) and (t,t+1) then, we can just do:</p>
<p>(y1-y2)=m(x1-x2) we get (t+1-0)=(1/4)(t-0) = 1/4=(t+1)/t;</p>
<p>solve for t we get -4/3</p>
<p>Thanks, the answer for the first one is A
anybody know how to solve the other 2?</p>
<p>^^ already posted number 2…</p>
<p>C) 3 should be the answer.</p>
<p>Long way:
Think about this logically. if its positive, must be above 0, and u can cancel out 1, 3, 5, 15. Because they are multiples of 15. If you think about the other numbers…12 works..goes into 15 1 time R3. 14, 13, 11, 10, 9, 8, 7 dont work. 6 works, 5 doesnt 4 works. and thats it…</p>
<p>4, 6 and 12 work. thats 3…</p>
<p>Short way:</p>
<p>If the remainder is 3, the “number” has to be 3 away from 15, and considering its division, it cant be 3 above 15…so it has to be 12. Now just concider the factors of 12… 1-12, 2-6, 3-4. Immediatly u can eliminate, 1, 3 because they go directly into 15. all u have left to eliminate is 2 because 2*7=14, remainder is 1, and ur left with 4, 6, 12</p>
<p>which is the other one??</p>