how do you do #46?
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 ((a-1)^2+(-3-2)^2)=13, and when you simplify it, you need to use the quadratic equation to find the x coordinates.
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).
I drew two lines connecting those points to (1,2) and labeled those segments with a distance of 13 because that was given in the question. Then I drew a vertical line from (1,2) to y=-3 or what would be the point (1,-3) and labeled that with a distance of 5 (the sum of the absolute values of y).
That vertical line makes a 90 degree angle with y=-3 and you already have two sides of both of the triangles made so you can use the pythagorean theorem to find the remaining sides (or the distance from (1, -3) to (x,-3) and (z,-3)).
Then all you need to do is take the value you recieved from the pythagorean theorem and add and then subtract it from 1 to find the x values of the two points.
I'm sorry if that was somewhat difficult to follow because it's hard to explain without drawing it but I know the pictures help me a lot more than using formulas!
I did it the same way Trisha did...
All it is, is the distance formula but instead of trying to figure out the distance you're GIVEN the distance and they want you to figure out a MISSING COORDINATE. so i set mine up the same way trisha did, 13^2=(1-X2)^2 + (2+3)^2 .... so you foil out the (1-X)(1-X) and after u clean it up it should look like 169= X^2 - 2x + 1 + 25
and then u just bring the 169 to the right side then u should have 0= X^2 -2x -143, and after that u just do the quadratic equation to figure out the 2 values for the missing X and pair them up with the given -3.... hope that helps clear things up
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).
Does #47 (1.2) just mean substitute (a,b) for x and y in the equation?
I think that's what it's saying. That's how I did it, at least.
well everyone is most likely asleep right now but i am stumped on number 59!
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.
For number 56, do you just plot all the graphs from the previous questions 53,54, and 55?
Im pretty sure thats what you do. Thats the way I did it.
Anyone understand 72?
for #55 how do you rotate it when its parallel to the origin?
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).
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
...like use two points to find the slope, and use one point, in the point-slope formula, to get to y=mx+b
I really hope that makes sense
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.
I might be wrong though? I think the equation would be something like a cubic function.
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.
If any of you are doing the 1.2 and 1.3 homework block, and your stuck on #72 there is a really easy way to do #72. Make a system of equations of the 3 points. For example, for (2,0), plug in 2 for x in ax^2+bx+c=y and 0 for y. Do the same thing for (4,0) and (0,1), and then solve the system.
can someone explain to me why the graphs of #71 are the same and different?
Hey guys how do you do number 76?
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!
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! :)
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.
Hopefully I did the right thing..(:
@Hey guys how do you do number 76?
Maaz, all you do is try and solve for "x".
1. Multiply the equation by "x" to get rid of the denominator of "x's" on the left side
2. You will get a+b=cx. Simply divide by "c" to get the "x" on its own
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.
The previous post was mine too, i forgot my last name...
Nancie my thought process , for number 71 for b : I said that the square root of x^2 is x, and the absolute value of x, is x.. so thats what helped show that they would have the same graph.
I just had a quick question on something we did in class a few days ago...
it had to do with
Could anyone explain #49? I don't get how to do it at all...
nick: for that problem you have to realize (in terms of variables) that f(x-h) -fx/ h is actually (fx-fh) - 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
Okay so for #49 you have to think of the outside border as a cylinder with a circular surface that isn't a complete circle. You know the volume of a cylinder is pi*r^2*h. The pi*r^2 is simply the area of the circular surface. So the new volume of cylinder is Ah=V, and we already know V and h, 27 cubic feet and 1/4 feet respectively. THen just plug them in the Ah=V to find the area of that ring. After that, find the area of the pool and add it to the area of the border to find the area of both pool and border. Then find the radius of this large circle by using pi*r^2=A.
Hope this helps to you and anyone else stumped.
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 circle-area 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 circle-the 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!
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!
# 44 is basically the same problem as #43 except instead of labeling each side of the original rectange x, you need to label the length as 2x since it is twice as long. (You may want to draw a picture). So your orignal rectangle's width is x and length is 2x. So then draw your 1ft boxes and write your equations for the remaining area. here is a picture (as long as the formating doesn't get messed up when I post this):
[ ]2x-2 [ ]
X [ ]
[ x-2 x-2 ]
[ 1 ] 2x-2[ 1 ]
The volume formula is L x W x D. They gave us the volume of 4 so just plug in your values. Length = 2x-2 Width = x-2 and depth = 1.
(2x-2)(x-2)(1) = 4
solve for x: one of your solutions should be extraneous.
Hope this helps!
sorry, looks like my picture didn't work. Just follow the example picture on page 44, and you should be good.
i have a question on #90 on the 1.6 homework. please help! :) thanks
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.
Now for the midpoint, use the midpoint formula to find the midpoint, and you'll see that both the x and the y coordinates are the same. So when u plug those into the y=x formula, it works and if you graph it, every point on y=x has the same x and the y coordinates. So you can prove that the midpoint of (a,b) and (b,a) lies on y=x.
^ thanks so much!
Does anyone know how to do #59 in section 1.4?
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.
can someone help me with # 90 in 1.5 the water bills problem.
So for this problem, you know that the $21.60 is the a mandatory price for the bill since it is a quarterly bill and the $21.60 is the fee for 1 quarter of year. Then, it says that for every 1000 gallons past 12000 gallons is $1.70. Therefore, you can set up the equation 21.60+1.70x. Now the question gives you an example homeowners bill whose price is ranged from $65.75 to $28.40. Well the equation made is to figure out the price of a bill, so set that equation equal to the two extremes of the range that we are given.
Hope it helps!
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.
Mrs. Johnson teaches math at Metea Valley High School in Aurora, IL.
Ap Ab Calculus
H Algebra 2 Trig
H Pre Calculus
Regular Pre Calculus