Could someone please show me how to start #66 in section 4.4? The word problems give me a lot of trouble.
Well that problem i understood the first part, so for part a) i just set the equation: -16x^2+96x greater than 112 because thats the time interval we're looking for (when it's more than 112ft)and then when you set it equal to zero, you have a quadratic. so then i used the quadratic formula and then you end up with x= 1.59sec and x= 4.41sec so then thats you're answer.
yea, and for part c and d, since this is an upside down parabola, the maximum could be found as the vertex. do (-b/2a) for the x-value (seconds), or you could just use the table function on your calculator. And obviously, once you find the vertex on the table, you'll ,know both x and y coordinates, which are seconds and feet respectively.
Can someone please show me how to do number 45 in section 4.4? This is as far as I got. (i know its less than or equall to 1 but i don't have that key on my computer)(x+4)/(x-2) < 1 (x+4)/(x-2) - (x-2)/(x-2) < 0(x+4)-(x-2)/(x-2) < 0
On number 45 in section 4.4 you have to first subtract 1 so the equation is less than or equal to zero then you find the critical points (zeros and vasys). The zero being -4 and the vasy being 2. Then set up a number line so the different intervals are easily seen and plug test numbers into the equation, I used test numbers 3 (and got 6), 0 (got -3), and -5 (got -6/7). For this problem your looking for any output value that is equal to or less than 0 and since x can't equal 2 (vasy) the solution to the problem is (-infinity, 2).
Could someone help me out with number 12 on the review problems? I'm confused with finding the bounds, when finding the zeros of a polynomial that does not have zeros for integers.
whoops!i meant integers for zeros, not zeros for integers.
Just like finding any other zeros you can use the rational root theorom. You will have to test possible zeros that you find while doing the rational root theorom.When you apply the rational root theorom to this problem you get plus or minus one, plus or minus 3.4 / plus or minus one, plus or minus 2, plus or minus 3, plus or minus 4, plus or minus 6, and plus or minus 12. After simplifying this you get plus or minus 1, plus or minus 3.4, plus or minus 1/2, plus or minus, 1/3, plus or minus, 1/4, plus or minus 1/4, plus or minus, 1/6, plus or minus 1/12, plus or minus 1.7, plus or minus, plus or minus 17/15, plus or minus 17/30, plus or minus plus or minus 17/20, plus or minus 17/60. The this is where the factor theorom comes in. You plug these possible zeros into the question. Your goal is plug it in until you reach 0 as your answer. After plugging in possible zeros -3.4 is one that works. After this, you do synthetic division or long division. For synthetic after plugging in -3.4 your remainder is zero. (Which means -3.4 is a possible zero!) The other numbers before that you use to create an equation. The equation would be 12x^2-1x-1. This technically means that 12x^3+39.8x^2-4.4x-3.4 divided by -3.4 = 12x^2-1x-1. Since, you already have 1 zero, you need 2 more zeros since the degree of the origional equation is 3. Using the quadratic formula (oppisite of by plus or minus square root of b^2-4ac all over 2a) with the equation 12x^2-1x-1 your other zeros come out to be -.33 and -.25. If you have any other further questions please let me know! I tried to be as detailed as possible.
I can't seem to get the answer for 44 that is on the syllabus answer sheet. I've figured out that the equation is x^2-7x+12<0 and the critical points are 0 (vasy) 3, and 4 (x-int). When I wrote out my different intervals (-infinity to 0, 0 to 3, etc) I got only 3<x<4 as an answer. Am I missing something or did I make some silly error?
Can anybody help me with #11 on the review problems? The only way I can get the zeros is on my calculator.
I am very confused on how to solve the equation f(x)=3(x^4)-4(x^3)+4(x^2)-4x+1. I know that then the leading term is (x^2), then you can use the quadratic formula, but I am not sure of how to solve it this way. If anyone knows, PLEASE EXPLAIN.To answer Alex Masiak's question (one above mine), you NEED your calculator to do it. Unless you can come up with some way to calculate it by finding zeros and all, then good job, but IF YOU SEE DECIMAL PLACES, then just be aware that you need your calculaor to solve for the zeros.
The problem I said above was #55 from the review problems in the homework. (sorry for the additional comment :))
Chandan-for your question use the rational root test first, and you will find that the equation has roots of 1 and 1/3. Then once you factor those out (synthetic division) you have a quadratic equation that you can solve.
ok thanks i got the part where you use the RRT and get the roots, but I seem to always get confused when it comes to the synthetic division. I'm confused as to WHICH ROOT you put in the dividend, in this case the 1 or the 1/3. If you divide both then you get two different quadratic equations, so what do you do? can you explain that a bit more in detail?
never mind Alex!! Sorry about that..I blanked out there. I got it now. I didn't really know how to approach it, and you telling me to use the rational root theorem really helped. Thanks! This blog really does help!
Either you can use synthetic division twice and use the ending formula obtained from the first and then use the zero as the divisor for the 2nd. or you can Multiply the 2 roots together and use long division for the polynomial
wait how do you do #11 on the review without a calculator? cause you kind of can't use the rational root theorem or the quadratic equation. is using the calculator the only way to solve it?
Chammy- yes the only way you can solve that is using the calculator...i don't really know of an actual way to solve it by hand.
I'm really confused on #65 from the review. Page 278. I understand that you can factor the top and the bottom and test numbers with the specific restrictions but the back of the book answer includes a 2 and a 6 and I'm not sure where they got that. If anyone could explain this I would be incredibly thankful!
Hey Lindsey :D 2 and 6 are the x-intercepts aka critical points.
does anyone understand to solve question 62 on the chapter review?the question asks to solve graphically and algebraically:(3-2x)/(2x+5) > (or equal to) 2
i know this is kind of last minute, but could someone help me with #92 from 4.1? its like the one we did in class yesterday and i'm still kind of confused with it. the question is what is the length of the edge of a cube if its volume could be double by an increase of 6 cm in one edge, and increase of 12 cm in a second edge, and a decrease of 4 cm in the third edge?
Hey Justin, the equation for volume would be V=x^3 because all of the side lengths are the same. Then, if you increased one by 6, it would become (x+6), if you increased it by 12, it would be (x+12), etc. You would end up with 2x^3=(x+6)(x+12)(x-4) because it says the volume would be doubled (and the original volume was x^3). Then you FOIL out all of the zeroes and get 2x^3=x^3+14x^2-288. Subtract 2x^3 (volume) from both sides to get 0=-x^3+14x^2-288. Use your calculator to find the zeroes, which should be 6, 12, and -4. -4 cannot be a length of the side, so 6 or 12 must be the two answers.
Could someone please help me with #67 in the section 4.4? I know it was not one of our homework problems, but when I was studying for the test I came across it and I was a little confused.
can someone please show me how to do problem #50 from section 4.4?5/(x-3)>3/(x+1)
Julianne,i hope this helps.....(also look at pg.270-pg.271-Example 3 "Solving a Rational Inequality")in order to solve problem 62 graphically, you just graph the equation and see when the equation has a y value greater than or equal to 2. Now to solve it algebraically, im not sure, so someone correct me if im wrong, but i think you set the equation equal to zero, then solve for the equations zeros and values of x where y is undefined. Then set up intervals with those points to plug test values into. Find when the values are positive and negative. where the values are positive is your answer to the problem.so....-(3-2X)/(2X+5)>= 2-set to zero-((3-2X)/(2X+5)) - ((2*(2X+5))/(2X+5)) >= 0 which gets boiled down to(-6X-7)/(2X+5)>=0 in this equationy = 0 at (-7/6,0)and a vasy at x = -2.5so you then plug in test values-inifinity < x < -7/6-7/6 < x < -2.5-2.5<x<inifinityfor what ever values that x > 0 (x is positive)is the answer to the problem.In this inequalitywhen x is....-negative inifinity < x < -7/6 y= positive-7/6 < x < -2.5 y = negative-2.5<x<inifinity y = negativenegative infinity < x < -7/6so when x isnegative infinity < x < -7/6y is > 0 in our equation set to zero when x = -7/6y = 0 in our equation set to zerowhich means that when x isnegative infinity < x < -7/6y > 2 in our original equation and when x = -7/6y = 2so when x isnegative infinity < x <= -7/6y >= 2 in our original equationso the answer isnegative infinity < x <= -7/6hope that helps.
o and if they wantedy <= 2youd want to know the x values wheny <= 0 (negative and equal to zero) in your equation set to zero - instead of when its positive
I have no idea if we're supposed to post here for chapter five questions but I didn't get problems 49 and 52 in tonight's homework. can someone please explain it? :)
I have the same question as Anika,i don't get #49, and # 43 I keep getting stuck.
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