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Sathvika Ashokkumar
8/29/2011 11:55:03 am
how do you do #46?
Trisha
8/29/2011 12:45:33 pm
I'm not sure if this is exactly right, but what I did was plug in the points (1,2) and (a, 3) equals 13 in the distance formula and solved with a quadratic equation. So basically, you have the square root of ((a1)^2+(32)^2)=13, and when you simplify it, you need to use the quadratic equation to find the x coordinates.
Brooke Eber
8/29/2011 01:18:32 pm
I'm not sure either, but I drew a picture and plotted (1,2) and the line Y=3. then I drew two points on the Y=3 and labeled them (x,3) and (z,3).
Rushad Marfatia
8/29/2011 01:38:39 pm
I did it the same way Trisha did...
Alyssa Andrade
8/29/2011 01:39:58 pm
I did the same thing as Trisha. I plugged the points (1,2) and (X,3) into the distance formula and set it equal to 13. After squaring both sides of the equation, you are left with a quadratic equation. Then by using the quadratic formula, you get the two solutions or answers to the problem (one is positive and one is negative).
Sally Kim
8/29/2011 02:07:33 pm
Does #47 (1.2) just mean substitute (a,b) for x and y in the equation?
Daniel Gilmore
8/29/2011 03:07:44 pm
I think that's what it's saying. That's how I did it, at least.
Nick Herrig
8/29/2011 03:08:11 pm
well everyone is most likely asleep right now but i am stumped on number 59!
Nick Herrig
8/29/2011 03:25:32 pm
nevermind :)
Nirav Surati
8/29/2011 03:28:18 pm
Aight, im going to show you how to do 59. First, you have to find the midpoint of a side using midpoint formula. Doesn't matter which one. After you find that point, use the distance formula with that point and the point where the median is coming from. For example, use distance formula on point A (0,0) with the midpoint coordinate of side CB. This will get you the lengths of the medians. Do this for all the sides and points, then just simplify each answer.
Alyssa Andrade
8/30/2011 10:52:38 am
For number 56, do you just plot all the graphs from the previous questions 53,54, and 55?
Ryan Aske
8/30/2011 12:54:00 pm
Im pretty sure thats what you do. Thats the way I did it.
Brooke Eber
8/30/2011 02:00:40 pm
Anyone understand 72?
jessica mantovani
8/30/2011 02:05:19 pm
for #55 how do you rotate it when its parallel to the origin?
Brooke Eber
8/30/2011 02:26:33 pm
If you mean 55 (1.2) you just make both the x and y coordinates negative. So if the original segment goes to (3,4) then the rotated one would go to (3,4).
Trisha Mahapatra
8/30/2011 02:37:46 pm
For #72, you can use any of the two points given, and create a y=mx+b equation...I think...because that's what I did
Trisha Mahapatra
8/30/2011 03:01:48 pm
...like use two points to find the slope, and use one point, in the pointslope formula, to get to y=mx+b
Andrea Lin
8/30/2011 04:49:18 pm
I'm not sure how #72 is done either, but I think all 3 intercepts have to be used, not just 2 of them. I don't think it would follow as a y=mx+b equation because the three intercepts don't look like they'd be a line since there are different slopes between all 3.
Daniel Gilmore
8/31/2011 05:52:32 am
I'm posting from the library with less than 30 minutes till the homework is due, so nobody will probably see this. Yet I'm posting it anyway. I did 72 with the equation of a cosine curve modified so that the period would be 2.
Mit Shah
8/31/2011 12:40:46 pm
Nancie Huynh
8/31/2011 01:44:44 pm
can someone explain to me why the graphs of #71 are the same and different?
Maaz Baig
8/31/2011 01:46:02 pm
Hey guys how do you do number 76?
Sally Kim
8/31/2011 02:18:07 pm
Nancie  y=abs(x) and y = sqrt(x^2) are equal because the y value will not change if x is negative or positive. For instance, say x = 4. for y=abs(x), that would mean y would equal 4 because of the absolute value. For y = sqrt (x^2), y would still equal 4 because x is squared before it's square rooted, so 4 becomes a positive 16 and THEN it is square rooted. (I hope this makes sense so far). On the other hand, y = x is just a linear equation. y = (sqrt x)^2, on the other hand, is different from y=sqrt(x^2) because in y=(sqrt x)^2, the x can not be negative because it's being square rooted first. Negative square roots = imaginary solutions, so y=(sqrt x)^2 IS NOT THE SAME THING AS y=sqrt(x^2). ONLY y=abs(x) and y=sqrt(x^2) are the same. If that confused you, I totally understand..it confused me the first time I was going through the problem too. Hope this helped!
Rose Thompson!
8/31/2011 02:27:47 pm
For #71, I just graphed it on the calculator and compared the domains. For b, they are the same because they both have a domain of x is greater than or equal to 0 and have a period of all reals. This is because the absolute value of anything is always positive, and any number squared is always positive. For c,the two graphs are different because y=square root (x)^2 has a domain of x is greater than or equal to 0 since you can't have a negative under the radical, while y=x is just a straight line with a domain of all reals. And you just do the same for d! Hope this helps a little! :)
Trisha Mahapatra
8/31/2011 02:34:09 pm
Maaz, for 76 I multiplied both sides by x, getting a+b=cx, and then I divided by c to get (a+b)/c=x.
Mike Tseng
8/31/2011 02:37:37 pm
@Hey guys how do you do number 76?
Stephanie
9/1/2011 04:36:30 am
for number 76 i also did the same thing as Michael, i divided by c to get x on its own, because the question was just basically asking to solve for x.
Stephanie Cuddalore
9/1/2011 04:41:21 am
The previous post was mine too, i forgot my last name...
Nick H
9/1/2011 03:03:12 pm
I just had a quick question on something we did in class a few days ago...
Sally Kim
9/2/2011 10:26:15 am
Could anyone explain #49? I don't get how to do it at all...
Nancie Huynh
9/3/2011 01:44:22 am
nick: for that problem you have to realize (in terms of variables) that f(xh) fx/ h is actually (fxfh)  fx/ h when you factor out the first part. The fx cancels out and you're left with fh/h. Then the h equals one so you are left with just h
Mit Shah
9/3/2011 03:21:46 am
@Sally
Rose Thompson!
9/3/2011 09:49:56 am
Hi! For #49, I did it a little differently: I called the pool combined with the border the outer circle, and just the pool the inner circle. Okay. So you know that the area of the border = area of the outer circlearea of the inner circle. The area of the inner circle is just pi(5)^2, and the area of the outer circle is just pi(5+x)^2, with "x" being the radius of the border (same as the drawing in the book). So if you subtract the outer circlethe inner circle, you get pi(10x+x^2) to be the area. Area*the height is the volume, which you already know to be 27 cubic feet, and you just found the area. You convert the height(3 inches)to feet (1/4 feet), and plug it in the equation for volume. Now your equation should look like pi(10x+x^2)*(1/4)=27. You simplify this and put it all on one side so you can do the quadratic formula. One of the solutions should be a negative extraneous solution while the other should be the correct answer. Hope this helps!
Alyssa Andrade
9/4/2011 08:13:59 am
Does anyone get how to do #44? I get how to do the problem before it when it's dealing with a square, but I can't figure it out as a rectangle. Please help!
Katie Uram
9/5/2011 08:45:49 am
@ alyssa
katie Uram
9/5/2011 08:46:55 am
sorry, looks like my picture didn't work. Just follow the example picture on page 44, and you should be good.
Rushad Marfatia
9/5/2011 01:40:58 pm
i have a question on #90 on the 1.6 homework. please help! :) thanks
Tirth Patel
9/6/2011 07:44:46 am
For 90, you need to find the slope of the points (a,b) and (b,a). Once you do that you'll see that the slope is 1. So, now you can prove that the line is perpendicular to y=x since the slopes are opposite reciprocals.
Rushad Marfatia
9/6/2011 12:59:01 pm
^ thanks so much!
Dylan Green
9/8/2011 12:29:43 pm
Does anyone know how to do #59 in section 1.4?
Nikki Beesetti
9/9/2011 03:39:11 am
For 59, how you want to set up the equation is 0.5x+1x=60 because they ask how much 12 karat gold should be mixed with pure gold to get 16 grams of karat gold and 12 karat is 1/2 half of pure gold. When you solve that you would then get x=40. and then substitute but then the thing is that in the book they say 40g of 12 karat gold should be mixed with 20g of pure gold but when you do that you get 40 as the answer but it should be 60. I think the book got it mixed up or something.
Rushad Marfatia
9/10/2011 08:23:39 am
can someone help me with # 90 in 1.5 the water bills problem.
Mit Shah
9/11/2011 04:06:57 am
#90
Mit Shah
9/11/2011 04:09:26 am
Also when you find the x values when you set the equation to each of the two extremes, add it to 12000 because that x value is the amount of gallons that is used after 12000 gallons.
Rushad Marfatia
9/11/2011 09:50:39 am
thanks brah! Comments are closed.

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