H Algebra 2 Trig Ch. 7

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H Algebra 2 Trig Ch. 7

11/18/2010

Hello Classmates! Feel free to post your questions and answers here. Teamwork is beautiful!

Archana Sathappan

11/19/2010 09:21:06 am

For the homework Pg 339 #3c, it asks for the removal discontinuity after graphing. I graphed it and I understand how to find the hole/removal discontinuity for only the x portion. How do I solve for the y portion of the hole/removal continuity. I got that the x hole is 2. In the back of the book it says that the removal discontinuity is (2,2). So, the y portion of the hole is 2 as well. But, how do you find that?

craig cook

11/21/2010 07:45:00 am

for #52 p.334 i cant seem to factor it even though it is probably somthing really simple

Nikita Hariharan

11/21/2010 08:35:31 am

#52.
answer: (3x-4)(2x+5)

You can solve this problem by trial and error. Just got to try it out. Hope this helps! :)

Johannes Grandin

11/21/2010 10:42:40 am

Just wondering how to factor number 19 on page 340.

Johannes Grandin

11/21/2010 10:45:26 am

For page 339 #3c, you can actually just plug it into the calculator and then look at your graph. Then you can go to your table and wherever it says ERROR for the y portion, that's the hole. The hole is also whatever you were able to cancel out when you factor the numerator and denominator.

Miranda Colin

11/21/2010 11:01:59 am

On page 340, how do you factor #53 and #61? For #61, I have a feeling it has something to do with the difference of two squares but I can't figure it out.

Johannes Grandin

11/22/2010 08:35:35 am

Hey I was just wondering if the quiz tomorrow is just on 7.3 and 7.4 or also on any other sections?

Ayo Adewole

11/22/2010 09:25:40 am

Can someone please do number 78
in section 5.3 please

Ayo

11/22/2010 09:27:55 am

Kirin Upadhyay

11/22/2010 09:48:40 am

for number 6 on page 360 would the answer to the problem be x cubed + ll x squared +40x because there was a 48 in the polynomial but since it was divided by 48 wouldn't that just be one?

Kirin Upadhyay

11/22/2010 09:50:06 am

oh, wait couldn't you just cancel out the 12 and the 4 in both equations then? yea nevermind

Nikita Hariharan

11/22/2010 09:55:56 am

Ayo:

I think you meant 7.3 so here goes...
78:

(p^5+q^2)(p^5-q^2)
This problem cannot be broken down further because of p^5 (not an even number). Hope this helps!

Ishta

11/22/2010 10:12:23 am

On P341, what is the best way to solve question number 47? Also, how would you solve question 58 on the P334?

Zach Zielke

11/22/2010 11:15:12 am

On page 339, how would you do number 17? I just need help with one and I can do the rest, thanks.

Christian Carvallo

11/22/2010 11:21:50 am

Christian Carvallo

11/22/2010 11:23:50 am

Zach the answer to 17 would be
factored out to be (4x+3)(3x+4) i solved it by guess and check. I'm not sure if there is a quicker way if there is someone please tell me because that took a long time to solve.

Ashley Bruner

11/22/2010 12:40:02 pm

How would you do #86 on page 333? I don't understand how you break down the square in this problem

Ritika Rao

11/22/2010 01:05:29 pm

Can somebody explain #85 on 333? I know it breaks down into [6-(x-5)][6+(6-5)] but I don't get how to get the "[6+(6-5)]" part of it.

Becky Wegner

11/22/2010 01:16:43 pm

Can someone help me with number 5 on tonights homework which was page 346?

also, i dont understand how you know when to change the sings

Aneesh Asokan

11/23/2010 09:33:41 am

When you do long division, how do you know when to make subtraction or addition in the answer??

Archana Sathappan

11/23/2010 12:29:11 pm

Aneesh- You just have to change the sign of every variable/constant of whatever the remainder is.

NIkita Hariharan

11/24/2010 01:39:02 am

Pg. 333
#85

36-(x-5)^2
(6+(x-5))(6-(x-5))
(1+x)(11-x)

Nikita Hariharan

11/24/2010 01:40:51 am

Pg. 333
#86

81-(2x+7)^2
(9+(2x+7))(9-(2x+7))
(16+2x)(16-2x)

Nikita Hariharan

11/24/2010 01:53:25 am

The answer for #5 on pg. 346 is x^2 -x+3.
I do not really know how to type out all the steps but I will try. So you start of with 2x cubed -3x squared +7x-3 . The first number you would plug in would be x squared so that the when multiplied by 2x-1 you would subtract 2x cubed- x squared. Your remainder would be -2x squared +7x. Then the next term you would plug in would be -x. When multiplied by 2x-1, you would receive -2x squared +x. Now, your remainder would be 6x-3. The third term you would plug in would be +3. When multiplied by 2x-1, you will receive 6x-3. Theis would help you get a remainder of 0. So the answer finally is: x squared-x+3. Hope this helps :)

Nikita Hariharan

11/24/2010 01:56:52 am

Oh and sorry Becky for the signs you always change the sign to whatever the constant/variable is. so for exaple, in the problem you asked, when there is a remainder of -2x squared +7, the first term is negative so the number you would plug in would have to get you to a negative that way when you subtract, the terms will cancel out.

Ishta

11/24/2010 02:07:15 am

Becky-the question is 2x^3-3x^2+7x-3/(2x-1). first take the 1st 2 terms, 2x^3-3x^2 and look at the 1st one. 2x^3 divided by 2x (from 2x-1)That would be x^2 so put that at the top of the radical. Then multiply x^2 and 2x-1. That should equal 2x^3+x^2. Then subtract it from the the orginal(2x^3-3x^2). DO NOT FORGET TO CHANGE SIGNS. That will you with -2x^2. Bring down the 7x. Now think again, -2x^2 divided by 2x=-x. Put negative above the radical. Multiply through and subract. You should be left with 6x. Bring down the -3. 6x divided by 2x is 3 so put that on top of radical. Multiply through and subtract. There should be no remainder. The answer is (2x-1)(x^2-x+3)=2x^3-3x^2+7x-3. Check answer by distributing and you are done.

Jelena Martinic

11/28/2010 05:02:37 am

I don't have my book with me and I’m not sure if I copied the problem wrong, but i can't seem to factor #54 on pg.333?

Emma Burck

11/28/2010 06:20:52 am

Jelena- The problem was 6x^2+37x-20.
I was able to factor it... try using 2x and 3x :)

Emily Cowgill

11/28/2010 02:04:07 pm

I can't seem to factor number 54 on pg. 334, could someone please help me out?
And I don't quite understand yet how to do number 82 on pg.334. I know we have done stuff like it in class, but I am still not quite getting it.

Hayley Vaughn

11/28/2010 03:18:20 pm

Does anyone know how to do number 59 on pg. 341 or number 1 on pg 346?
I got through most of the #1 on pg 346 but I got stuck.

Ryan Vasicek

11/29/2010 02:17:55 am

@ Emily

number 54 is (3x + 20) (2x - 1). it just takes some guess and check

Tucker Jacob

11/29/2010 07:15:54 am

i do not understand how to divide number 13 on pg 346. Can anyone explain and show me how to do it step by step

thanks

Archana Sathappan

11/29/2010 09:30:43 am

The problem for 13 on pg. 346 is x^3+3x^2+3x+1/x^2+2x+1. First do x^3/x^2 to get x and place that right on top of the x portion of your division set up. Then multiply all the terms (x^2+2x+1) by x to get x^3-2x^2-x. Then change the sign of all of those that you can take the difference of that from x^3+3x^2+3x+1. The remainder is then x^2+2x+1. Then you do x^2/x^2 to get 1 and you place that right on top of the constant portion of your divisiom set up. Then multiply 1 by x^2+2x+1 to get x^2+2x+1. Then change the sign of all of the terms so that you can take the difference of that from x^2+2x+1. The remainder is 0. At the top in your "answer section" you have x+1 as your answer and there was no remainder, so the answer is x+1. Hope this helps!

michelle filipek

11/29/2010 01:14:17 pm

how do you do #39 on page 340?

michelle filipek

11/29/2010 03:08:33 pm

For our homework on page 346 should number 17 have a remainder? and how would you do number 19?

Jacqueline Pedersen

11/30/2010 02:22:46 am

For number 17 on page 346, yes you should end up with a remainder of 2. So your answer would end up like:
x^3+x^2-x-2-(2/x+6)

Jacqueline Pedersen

11/30/2010 02:33:43 am

For number 19:
You have x^4-1/x-1
first you find out what times x=x^4, which would be x^3.
Then you times x^3 by x-1 so you get x^4-x^3.
Subtract the x^4 and bring down the -x^3 and -1
Figure out what times x=x^3, which is x^2
Times x^2 by x-1 so you end up with x^3+x^2, subtract then bring the x^2 and -1 down
Do the same thing for x^2- 1 and after subracting you should end up with -x+1
Figure out what goes into that and your final answer should come out to be:

x^3+x^2+x+1

Jacqueline Pedersen

11/30/2010 02:41:11 am

#39
It is a perfect square
So you find the square root of 4x^2< which is 2x. Do the same thing for 9y^2, its root will end up to be 3y. So you have (2x_3y)^2. In oder to find what sign, you look at the +9y^2, which means you would need to put a minus sign in order to end up with a postive 9y^2. You would end up with:

(2x-3y)^2

Jelena Martinich

11/30/2010 08:05:59 am

Can someone please explain how to solve tonights problems on page 353? I just need one example 'cus once again - i left my book at home and only copied the prbolems :/

Nikita Hariharan

11/30/2010 08:27:32 am

Jelena:
it's from today's notes.
for ex.
2x^3-3x^2-11x+6=f(x)
plus or minus (factors of your constant over factors of the leading coefficient)
so in this case it would be (1,2,3,6, over 1,2) (these are coming from the first and last numbers) = plus or minus (1, 1/2, 2, 3, 3/2, 6) these came from putting all the constant over the coefficients. then you have to plug these back into the originial equation. So for example f(3)= 2(3)^3-3(3)^2-11(3)+6= 54-27-33+6+ 27-33+6=33-33=0. So now we know that x=3 is one of our factors so we have (x-3)and continue this same process with all of the numbers ( you should end up with 3 factors) and will get (2x-1) and (x+2)!
Hope this helps :)

Aneesh Asokan

11/30/2010 09:09:44 am

I just want to make sure, most of the answers for tonights homework i got to only two answers. Not like the one we did in class today. Just want to know if I am going the right way, or if i totally messed up or anything.

Emily Cowgill

11/30/2010 11:43:59 am

Aneesh - I actually got 2 or 3 answers for the ones I found answers to. I'm not sure if I did it right though.

I know this is old stuff, but I still don't quite get how to do problems like number 37 on pg. 353. Can anyone explain it to me?

Emily Cowgill

11/30/2010 11:47:03 am

previous cont.

I tried the GCF thing, but could I take just an exponent as a GCF when they do not have the same base?

Meltem Cazimi

11/30/2010 11:56:14 am

emily! uhmm no you cant just take an exponent out as a GCF it has to have the same base, you need to look at what you cubed to get the number thats there. for number 37 you have to cube r to get r^3 and you have to cube s^2 to get s^6, so then you do (r-s^2)(r^2+rs^2+s^4) because you take and square the first and second numbers in the first perenthesis to get the first and last number in the second and then just multiply those two to get the middle. i hope that makes sense.

Meltem Cazimi

11/30/2010 11:59:05 am

for the first two problems on the homework, numbers 3 and 7, i plugged the equations in the calculater and the graph shows both of them crossing the x axis which means theres a solution, but in the table there are no 0's to indicate which would be solutions, also when i used the factor theroem i couldnt figure out what would be the solutions, help ?

Archana Sathappan

11/30/2010 12:28:33 pm

The way you check to see how many answer you will get is by looking at the highest exponent in the problem. SO if you had something like 2x^3+2x^2+5+1, you would have 3 answers b/c the highest exponent is 3. Hope this helps Aneesh.

craig cook

11/30/2010 01:27:32 pm

Meltem, even though the graph touches the x-axis, the solutions are not integers so that is why none of the numbers equal 0, so u r right about there being no solution. About the other question, when you find which number(s) are equal to 0, take those number(s), flip the sign in front of them and put and x in front of it. Here is an example:
#3 – the factors of 66 are 1,2,3,6,11,22,33,66. the factor of 1 is 1. when you work it out, 11 and -3 both equal 0, so you switch the sign and put an x in front. You end up with
(x-11) or (x+3) both of these would be appropriate. You can use either of those to do long division on the main equation. You would be doing x^3-10x^2-17x+66 divided by either (x-11) or (x+3). Hope it helps.

Jordan Crouch

11/30/2010 02:13:49 pm

Alright so on pg 346, number 9, how would you factor that? I got halfway through it and then got lost.

Brandon Jachimiec

12/1/2010 06:41:35 am

@Jordan Crouch
If you got halfway through then lost, you probably made a simple mistake. To factor you do x^2 times x to get x^3 which you put under x^3. Then you do x^2 times 5 to get 5x^2 which goes under the 7x^2. Then you flip the signs for the x^3 and 5x^2 (which are underneath) in this case, both become negative and you subtract to get 2x^2 left over. From there, you multiply x by 2x to get 2x^2. Then you do 2x times 5 to get 10x, flip both of the signs underneath (making them negative), and subtract leaving you with -10x and you bring down the -49. Again you multiply x by -10 to get -10x. Then you multiply 5 by -10 to get -50. Flip both signs, this time they become positive,add, and you are left with a remainder of 1 which you just put over x+5 (what you are dividing everything by.

This makes your final answer
x^2 +2x -10 + (1/x+5)
Hope I helped

Kayla Mann

12/1/2010 10:16:56 am

For last night's homework on page 353, does anyone know how to solve number 15?

Aneesh Asokan

12/1/2010 11:20:17 am

Hey, I had a quick question on the Homework on page 360 3-36 by 3s. I don;t get how to do any of them. I Sort of get it but I sorta Don't get it too.

Tyler Yanisch

12/1/2010 11:25:00 am

On page 353, how do solve number
19?

Karthik Makayee

12/1/2010 12:07:26 pm

@Kayla
Alright so u start off with the formula johnson gave us, which is: +-(factors of constant/factors of leading coeficient). the factors of constant is the last number in the equation and the leading coefficient is the first #. So once youve found the factors, u end up with: +-(1,2/1,2) - now u divide the numbers in order - ex)1/1, 1/2 , 2/1, 2/2 (u dont write down repeating #'s), so now u have: +-(1,.5,2). So next u can either plug that into your calc. or plug the 3 #'s to the original equation. If it equals 0, then it's a factor! Once you've done it by hand or your calc. you'll find out that .5,-1,-2 are all factors because they equal 0 when they're plugged into the original equation. Finally, u put a x infront of it and change the sign. And you end up with (x-1/2)(x+1)(x+2), and "x-1/2" is the same as 2x-1.

@Tyler
Just like the problem Kayla asked, u follow the same steps, u use the equaton: +-(factors of constant/factors of leading coeficient), and once youve found the possible factors. I simply plugged them into my calculator and found that (.5,-.5,-1/3) equal 0. So now i simply attach an X and change the sign, and you have, (x-.5)(x+.5)(x+1/3) which is the same as (2x-1)(2x+1)(3x+1).

Ishta

12/1/2010 12:14:21 pm

On page 360, how would you do number 33? Thanks :)

Karthik Makayee

12/1/2010 01:01:05 pm

@Aneesh
Ok lets start with #3, if u look at pg358 and at the very bottom it talks about "special cases" there are 3 cases. The 1st, is if the numerator and denominator are teh same, it equals 1. The 2nd, if the num. and den. are both subtraction and the denominator is switched, it equals -1. Finally the 3rd case, if the numerator and den. have different symbols or conjugates, they cannot be reduced any further. So these cases are used on problem 3) a=-1 b/c the num and denominator is subtraction and switched, b=1 b/c the num/den are equal, c=cannot be simplified any more, d=1 b/c the num and den are equal like in b.

Now that we kno how 2 simplify, the next Q are just simple math with factoring. I'll explain #6,7. 6) So u have (x^2+7x+12/12) * (4/x+4), u begin by factoring "x^2+7x+12" and you get (x+3)(x+4). Now u equation is ((X+3)(x+4)/12)*(4/(x+4)), so just like regular multiplication, u can cross off common factors. You cross off "x+4" on both sides, leaving ((x+3)/12)*4), then u can simplify 12 and 4 further. Leaving ((x+3)/3) as you final answer.

For #7 u do a very similar step, when your solving a division problem, u begin by flipping (x+1)/(x+5), thus making the problem multiplication, and u follow through. You factor out, and cross off (x+1), then multiply across. Getting ((x+3)(x+5)/5x), u then multiply (x+3)(x+5) together and get: ((x^2+8x+15)/5x) as your final answer.

craig cook

12/2/2010 08:40:12 am

would anyone know how to do #33 on our 7-7 homework?

Archana Sathappan

12/2/2010 09:03:47 am

@ Craig:
The problem for 33 on 7.7 is 3x^2-2xy+6x-4y/3x^2+xy-2y^2 * x-2/x^2-4. For the 3x^2-2xy+6x-4y part you can simplify the way Mrs. Johnson taught us a while ago. Take the first 2 terms from that portion and gcf it and then take the last 2 terms of that portion and gcf it. So you get x(3x-2y)+2(3x-2y) and that can be written as (x+2)(3x-2y). For the x-2 part from the original problem, you can not simplify it, so leave it as is. For 3x^2+xy-2y^2 you can factor that to get (x+y)(3x-2y). For the x^2-4 part, you can see that it is a difference of two squares, so you can simplify that to (x+2)(x-2). Now when you multiply across you get (x+2)(3x-2y)(x-2)/(x+y)(3x-2y)(x+2)(x-2). Cross out all like terms, and your final answer will be 1/x+y. Hope this helps.

Rohan

12/2/2010 09:42:42 am

Hey, Can someone just show me the steps me ro problem 27 on page 360 for the 7.7 hw? im gettin the right answer I just dont think I am doing it right.

Amir Khalid Richardson

12/2/2010 10:10:01 am

Rohan to solve for number 27 you should factor it so it is (x+5)(x+8)/(5-x)(x+8) after you cancel out the eights you should end up with (x+5)/(5-x), which can't be factored

Christian Carvallo

12/2/2010 11:16:23 am

Hey guys I need some help on number
3 on page 360. I know it looks pretty easy but for some reason i can not figure it out i am really stuck right now for some reason.

Aneesh Asokan

12/2/2010 11:25:36 am

Hey,Number 33 on Tonight's homework is super confusing. Can someone, by any chance, help me out.

Archana Sathappan

12/2/2010 11:31:15 am

@ Christian
For part a of problem 3 on pg 360, the question is x-5/5-x. You need to factor out a -1 from the bottom so that you gett x-5/-1(x-5). Then you can cancel the like terms and get -1 as your answer.
For part b it is x+5/5+x which is just 1, you can use the communative property to flip x+5 to 5+x, so you cancel both of the terms on numerator and demoniator b/c they are the same and the answer is 1.
For part c it is x-5/x+5. You can't do anything with this, so your answer is just x-5/x+5.
For part d it is x-5/x-5. Cancel both terms to get 1.
Hope this helps!

Archana Sathappan

12/2/2010 11:34:03 am

@ Aneesh
I have already answer 33 on pg. 360 for Craig above, you can look at that to see if it helps.

Christian Carvallo

12/2/2010 11:39:00 am

Thank you archana

Dan Casey

12/2/2010 11:59:40 am

Pg. 360 #30, does the sum of the two cubes cancel out the (x^2 -2x +4)?

Zain

12/2/2010 12:16:26 pm

What is the answer to number 12 on page 360?

Archana Sathappan

12/2/2010 01:01:16 pm

@ Zain
For the answer for 12 on pg 360, I got -x(x+3)(x-1).

@ Dan
After you split up x^3+8, you get (x+2)(x^2-2x+4). Since there's another (x^2-2x+4) in the denominator, that portion will cancel out. So, you question was if they cancel out, and it would be Yes.

Michelle filipek

12/3/2010 05:30:01 am

how would you do #33 on page 360?

Archana Sathappan

12/3/2010 08:56:29 am

@ Michelle
I have already answered 33 on pg 360 for Craig above. You can see if that helps you.

Johannes Grandin

12/4/2010 05:10:07 am

How would you do number 44 on page 362. It is giving me some trouble.

Ishta

12/4/2010 08:28:51 am

Thanks Archana for answering my question.

Noelle Linden

12/5/2010 04:41:30 am

Hey, i was wondering if someone could explain how to do number 48 on page 362 for the 7.7 homework?

Ashley Bruner

12/5/2010 07:43:30 am

Noelle-
For #48 the exaple above it in the book really helps. You would multiply the entire equation by the the minor denominators (the denominators of the fractions within the fraction itself). The denominators would cancel out but you're left with the minor denominator of the oppisite fraction. You multiply that with the equation (12-6 or 12-3) and then simplify. I hope that helps! And i'm sorry if i did it wrong

Miranda Colin

12/5/2010 10:46:00 am

Hold up, sorry Ashley but I'm still kind of confused on #48. In your explanation you canceled out the minor denominators right away but the example in the book didn't. Did someone else do it slightly different than Ashley? And can someone give me an answer to #48 because I have an answer but I'm really unsure if it's right.

raj shah

12/5/2010 01:05:56 pm

okay so for the new way
of solving by dividing
how do u know what number
to put in
the ledge
if u they dont give you a
factor..

Ishta

12/6/2010 09:07:34 am

Does anybody know how to do #30 on page 368? Just tell me what I need to multiply.
Thank You

Ishta link

12/6/2010 09:10:46 am

Raj- you use the factor theorem, and find one factor, then put it in the ledge and solve to find the other(s).

Ashley Bruner

12/6/2010 09:59:35 am

Could somebody please show me how to do #s 30 and 33 on page 369 from tonights homework?

Melina Kanji

12/6/2010 12:41:59 pm

Can somebody help me with question 18, 30, and 33 from tonight's homework, please?

Johannes Grandin

12/7/2010 02:53:00 am

Ishta, Ashley, and Melina:

For # 18 on yesterday night's homework you know that 9a^2-4a^2 is a difference of two squares,(3a-2b)(3a+2b). To get the common denominator, just multiply the top and bottom of the second fraction by 3a-2b. You do not need to multiply the 1st fraction because its denominator is already the least common one (3a-2b)(3a+2b). Then you just subtract whatever you get on top for the second fraction from 3a (the numerator of the 1st fraction), to get your answer. Make sure to cancel out after doing the subtraction if possible to get your final answer.

For # 30,you first square both the top and bottom of the first fraction to get ((x+y)^2)over(x-y)^2). To make it easier to understand, write the second fraction as 1 OVER 1 instead of just 1. Then, just multiply the numerator and the denominator of the second fraction by ((x-y)^2) to get your common denominator. Then do your operation by subtracting the second fraction's numerator, which is ((x-y)^2) from the first fractions numerator, which is ((x+y)^2). Remember that it's easier to write everything out in the completely factored/expanded form because you can then cancel out everything much easier if it's possible.

Here we go, one more. For # 33, it looks scary, but it's actually not that bad. Since you have three fractions with different denominators, you can multiply each denominator by the other two: the first fraction's denominator is (x-1), so multiply it by (x-2) and (x-3). Multiply the second fraction by (x-1) and (x-3) (on top and bottom) and the third fraction (again top AND bottom) by (x-1) and (x-2). When you have done all of this and have your fractions complete with common denominators and written out in expanded form, then and only then do your operations.

By the way (x^2) means (x-squared)and so on, I just didn't know how to do superscripts, so sorry if it's a little confusing.

Archana Sathappan

12/7/2010 10:59:05 am

Can someone explain how to graph 22 on pg. 376 besides plugging it into the calculator?

Can someone explain 27 on pg 376? I get that you need to first factor the denominators and then get a common denominator, but I'm kinda getting stuck.

#31 on pg 376 seems to be easy, but I can't seem to get it. Can someone help me?

thanks in advance

Zach Zielke

12/8/2010 08:58:26 am

I do not get at all how to graph any of the problems from 7.9 homework. I've been plugging them into my calculator and taking those graphs but if we can't use our calculators can someone show me how to graph them?

thanks

Ryan

12/8/2010 01:43:28 pm

Ryan

12/8/2010 01:46:44 pm

I don't know how to do number 6 on page 376. I got to x-3/x+ 1 with a hole of -2 but I don't know what to do from there. HELP!!

Christian Carvallo

12/8/2010 02:20:58 pm

Ryan, for number 6 the coordinates of the hole would be (-2,-5) you can find that by replacing x with -2 to find the y coordinate. The VASY would be at -1 because if you substitute -1 of x you get zero in the denominator. And the HASY would be at y=1 because both the numerator and denominator have high powers of 2 so by taking the highest coefficient of the two, (they are both 1) you get 1. hope this helps.

Johannes Grandin

12/9/2010 02:42:18 am

Hey I was just wondering if we have gone over 7-10 already or if we are doing that today (Thursday) in class. I think that's the homework we were supposed to do for today but I'm not sure. Does anyone know???

craig cook

12/9/2010 07:56:53 am

would anyone know how to do #30 on page 381?

Erin Hohman

12/9/2010 10:17:34 am

Can anyone help me out with question #30 from pg.381..i dont understand how to find the x's?

Emily Cowgill

12/11/2010 04:50:34 am

Does anyone know how to reset a TI-84 Plus silver edition? i have been trying to do the finals packet and when i plugged in the matrix to find the determinant, it said "error: invalid dim". does anyone know what that means?

Kristina C.

12/11/2010 05:42:42 am

Emily-
to reset your calculator I believe you click the [2nd] button on the calculator, then you click the addition sign [+], then the number "7", then "1", then "2". Hopefully this works!

Zain

12/12/2010 05:03:14 am

What's the equation for number 2 on page 393?

Ashley Bruner

12/12/2010 06:48:40 am

Could someone please tell me how y varies with x in #14 on page 397? I can get the rest of the problem if i know that

Miranda Colin

12/12/2010 10:49:08 am

Ashley, I don't have my book with me because I finished the homework in school so I don't remember what the problem was about. However, from what I understand, if y increases while x decreases, or vice versa, then it's inversely. If both y and x increase or both y and x decrease then it varies directly.I'm not entirely sure if that's what you were asking for so sorry if I didn't answer your question.

Jake Bennema

12/12/2010 12:35:35 pm

FOr #14 on page 397, I am pretty sure for question a, it is an inverse proportion, but I am not sure how to set up the general equation, any help would be great. thanks

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H Algebra 2 Trig Ch. 7

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