<p>I have a couple questions on the following blue book questions:</p>
<p>-page 373 #2 </p>
<p>"A rectangle with a length of l inches and width of w inches,a perimeter of p inches, and an area of 36 square inches. If l and w are integers, what is one possible value of p?"</p>
<p>I figured that: w*l=36 w+l=p; w could = 3 and length could = 12; so 3+12 = 15. The possible answers though are 24, 26, 30, 40 or 74. What am I doing wrong?</p>
<p>-page 373 #5</p>
<p>"Tim wrote a seven-digit phone number on a piece of paper. He later tore the paper accidentally and the last two digits were lost. What is the maximum number of arrangements of two digits, using the digits 0 through 9, that he could use in attempting to reconstruct the correct phone number</p>
<p>Since the area = length X width = 36. There is only a few combinations of integers that can make this area
You can do 1 X 36 , 2 X 18 , 3 X 12 , 4 X 9, and 6 x 6. Now just calculate the perimeter for any one of them.</p>
<p>For 1 x 36. p = 74. </p>
<p>2 X 18, p = 40 </p>
<p>3 X 12, p = 30</p>
<p>4 X 9, p = 26</p>
<p>6 x 6 , p = 24. There you go.</p>
<p>For the next one.</p>
<p>Its basically saying how many combinations can you get with the numbers 0 through 9. Order does matter since this is a telephone number. With 0 as the lead coefficeint you can have
0 0
0 1
0 2
0 3
0 4
0 5
0 6
0 7
0 8
0 9</p>
<p>So you can 10 combinations when 0 is the lead. Well you will get 10 more when 1 is the lead, 2 is the lead, 3 is the lead, 4 is the lead, 5 is the lead, 6 is the lead, 7 is the lead, 8 is the lead, and 9 is the lead. Add them all up and you get 100.</p>
<p>This problem is about… permutations, I think. What it’s basically asking you is how many permutations (combinations in non-math speak) are possible for a 2-number string, containing numbers 0 through 9.</p>
<p>It’s a bit like probability, in that you multiply the possibles together. Think of it this way:</p>
<p>_ _</p>
<p>Those two lines indicate two possible numbers.</p>
<p>10 10</p>
<p>What I did here is to fill in how many numbers can possibly fill that slot. Since in this problem, numbers aren’t “used”, you can reuse numbers (otherwise I’d write 10, 9). Just multiply the two numbers and you get 100, which is the answer.</p>
<p>You can use this principle to solve any combination/permutation problem; just draw the slots, fill in how many numbers are possible, and multiply them.</p>
<p>Example: Bob has six objects. He picks them one at a time randomly out of an abnormally large hat, uses them, and then throws them away. In how many ways can he pick them all?.
6 5 4 3 2 1
6x5x4x3x2x1=720. Bob can pick the objects in 720 different ways.</p>
<p>Hope I helped; this is my first post here so I’m not quite familiar with how things are done around here…</p>
<p>yeah you did not take into account perimeter (p) means 2w+2l. Instead you just took the length (12) and width (3) and added those up. That only constitutes to half the perimeter.</p>