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?
for #52 p.334 i cant seem to factor it even though it is probably somthing really simple
#52.answer: (3x-4)(2x+5)You can solve this problem by trial and error. Just got to try it out. Hope this helps! :)
Just wondering how to factor number 19 on page 340.
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.
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.
Hey I was just wondering if the quiz tomorrow is just on 7.3 and 7.4 or also on any other sections?
Can someone please do number 78in section 5.3 please
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?
oh, wait couldn't you just cancel out the 12 and the 4 in both equations then? yea nevermind
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!
On P341, what is the best way to solve question number 47? Also, how would you solve question 58 on the P334?
On page 339, how would you do number 17? I just need help with one and I can do the rest, thanks.
Zach the answer to 17 would befactored 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.
How would you do #86 on page 333? I don't understand how you break down the square in this problem
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.
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
When you do long division, how do you know when to make subtraction or addition in the answer??
Aneesh- You just have to change the sign of every variable/constant of whatever the remainder is.
Pg. 333 #8536-(x-5)^2(6+(x-5))(6-(x-5))(1+x)(11-x)
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 :)
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.
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.
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?
Jelena- The problem was 6x^2+37x-20. I was able to factor it... try using 2x and 3x :)
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.
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.
@ Emilynumber 54 is (3x + 20) (2x - 1). it just takes some guess and check
i do not understand how to divide number 13 on pg 346. Can anyone explain and show me how to do it step by stepthanks
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!
how do you do #39 on page 340?
For our homework on page 346 should number 17 have a remainder? and how would you do number 19?
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)
For number 19: You have x^4-1/x-1first 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 -1Figure out what times x=x^3, which is x^2Times x^2 by x-1 so you end up with x^3+x^2, subtract then bring the x^2 and -1 downDo the same thing for x^2- 1 and after subracting you should end up with -x+1Figure out what goes into that and your final answer should come out to be:x^3+x^2+x+1
#39It is a perfect squareSo 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
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 :/
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 :)
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.
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?
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?
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.
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 ?
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.
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.
Alright so on pg 346, number 9, how would you factor that? I got halfway through it and then got lost.
@Jordan CrouchIf 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 answerx^2 +2x -10 + (1/x+5)Hope I helped
For last night's homework on page 353, does anyone know how to solve number 15?
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.
On page 353, how do solve number19?
@KaylaAlright 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.@TylerJust 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).
On page 360, how would you do number 33? Thanks :)
@AneeshOk 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.
would anyone know how to do #33 on our 7-7 homework?
@ 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.
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.
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
Hey guys I need some help on number3 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.
Hey,Number 33 on Tonight's homework is super confusing. Can someone, by any chance, help me out.
@ ChristianFor 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!
@ AneeshI have already answer 33 on pg. 360 for Craig above, you can look at that to see if it helps.
Thank you archana
Pg. 360 #30, does the sum of the two cubes cancel out the (x^2 -2x +4)?
What is the answer to number 12 on page 360?
@ ZainFor the answer for 12 on pg 360, I got -x(x+3)(x-1). @ DanAfter 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.
how would you do #33 on page 360?
@ MichelleI have already answered 33 on pg 360 for Craig above. You can see if that helps you.
How would you do number 44 on page 362. It is giving me some trouble.
Thanks Archana for answering my question.
Hey, i was wondering if someone could explain how to do number 48 on page 362 for the 7.7 homework?
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
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.
okay so for the new wayof solving by dividinghow do u know what number to put in the ledgeif u they dont give you a factor..
Does anybody know how to do #30 on page 368? Just tell me what I need to multiply.Thank You
Raj- you use the factor theorem, and find one factor, then put it in the ledge and solve to find the other(s).
Could somebody please show me how to do #s 30 and 33 on page 369 from tonights homework?
Can somebody help me with question 18, 30, and 33 from tonight's homework, please?
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.
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
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
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!!
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.
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???
would anyone know how to do #30 on page 381?
Can anyone help me out with question #30 from pg.381..i dont understand how to find the x's?
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?
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!
What's the equation for number 2 on page 393?
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
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.
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
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