<p>What is the coordinate of the point on a number line that is exactly halfway between the points with coordinates 53 and 62?</p>
<h1>12 Page 475:</h1>
<p>If x squared - y squared = 77 and x+y=11, what is the value of x?</p>
<h1>10 Page 489:</h1>
<p>Stacy noted that she is both the 12th tallest and the 12th shortest student in her class. If everyone in the class is of a different height, how many students are in the class?</p>
<p>a. 22
b. 23
c. 24
d. 25
e. 34</p>
<h1>16 Page 491:</h1>
<p>If x is an interger greater than 1 and if y=x+1/x, which of the following must be true?</p>
<p>i. y =/= x
ii. y is an interger
iii. xy>x squared</p>
<p>a. i only
b. iii only
c. i and ii only
d. i and iii. only
e. i, ii, and iii</p>
<p>For 10) the answer is 57.5<br>
All you have to do is take (53 + 62)/2 = x
That’s really simple math, I mean all your doing is finding the average (mean) of the two numbers…</p>
<p>For number 12) the answer is (x=9)
You just have to know how to deal with simultaneous equations:
First, you need to solve for one variable in terms of the other
You have x + y = 11
So x = 11 – y
Then plug x=11 – y into the equation x^2 – y^2 = 77
So (11-y)(11-y)-y^2 = 77
So 121 - 22y + y^2 - y^2 = 77
So 121 - 22y = 77
So -22y - -44
So 22y = 44
So y = 2
Plugging y=2 back into the original equation x + y = 11 We can see that x = 9</p>
<p>I used the exact same methods as spratleyj post #2 for the first two problems. </p>
<p>For #10 on p. 289, it’s 23 because you have say the 12 shortest students and the girl is the tallest of them. Now to have her be the 12th shortest there are 11 people taller than her. </p>
<p>For #16 p. 491, try out the three things that may or may not be true.
x>1
y = x+1/x, which I’m guessing means x+(1/x) rather than (x+1)/x. Is that right?</p>
<p>i. Because I’m assuming y = x+(1/x), y≠x because if y=x then (1/x) would have to be zero. (1/x) is definitely not zero because the numerator is nonzero and 1/x is defined too since x≠0. </p>
<p>Another way to check this is to assume y=x and use that in the equation
If y=x+(1/x) and y=x, then
x = x+(1/x) by replacing y with x
0 = 1/x
0=1, a falsehood so the assumption that y=x is false, meaning y≠x is true. So the statement is true. </p>
<p>ii. y doesn’t have to be an integer. If x=3, y=3+(1/3). This is false. </p>
<p>iii. xy>x^2? Just plug in what you know.
xy = x*(x+(1/x)) = x^2+x/x = x^2+1
Does x^2+1>x^2 have to be true? Yes, it’s one greater no matter what. To really see it, subtract x^2 on both sides to get 1>0. Is 1>0 always true? Yes, so this is true. </p>
<p>■■■■ I feel like a moron now. I had 16 wrong because I forgot that it had =/= and not =. I read it as y = x and thought, hey that can’t be right because x is always added to 1/x which would make it more than y so they couldn’t be equal. Like if x was 4, y would be 4 + 1/4. y and x could never equal. I knew iii was true so I marked b but now I realize the careless mistake I made.</p>
<p>kthnx man and you were right to assume that it was x + (1/x)</p>
<p>Yoda, yeah when I was writing the post, somewhere I had “y≠x is true so (i) is false,” but then I was like, “wait, what was it asking?” It might be good to check what the problem was saying before working it out, then checking it after as well.</p>