5

13. Solve for x: and write the answer in interval notation:-1>30Divide using Algebraic Long Division: (2x' 3x+ 1) + ( -3)Divide using synthetic division: (5...

Question

13. Solve for x: and write the answer in interval notation:-1>30Divide using Algebraic Long Division: (2x' 3x+ 1) + ( -3)Divide using synthetic division: (5x+ [0x' 16x + 22) +6+2)

13. Solve for x: and write the answer in interval notation: -1>30 Divide using Algebraic Long Division: (2x' 3x+ 1) + ( -3) Divide using synthetic division: (5x+ [0x' 16x + 22) +6+2)



Answers

Solve each rational inequality. Give the solution set in interval notation. $\frac{1}{x+2}>\frac{1}{x-3}$

We'll start by distributing the three into the prince. He's on the right side, so three times and X is three x and three times a minus. Two is of minus six. When it's attract three x softball size, it's three exits will cancel once have one is less than negative six, which is not true. So this means there are no solutions. So we have no in a real notation on this.

So let's multiply every term by 15 s so that we can clear these fractions and then we'll solve, simplify or solve this inequality. So it's mostly by 15 in every term, cause 12 actually was not not want. 15 12 15 and three all go into, Let's see here not 39 45 maybe 60 60. So small time by 60 on every return here. So $60. 12 is five. So five times X plus three 60 divided by 15 is four. So let's do four times the X minus five and then 62 of the three is 20 so we'll have 22 times 20. Distribute the five minutes of five times access five X 5 10 Street is the plus 15 and then the four of well, four times X is for X and four times of minus five is minus 20 two toes Twenties 40 and we'll solve the inequality. Put the five accident forex together to make Nynex and 15 minus 2 20 isn't minus five. When you add five to both sides, you get nine X is less than 45 then divide by If nine on both sides. You get X is less than five. So, really, we're looking at a negative infinity all the way up to but not including five.

In this problem were asked to solve the rational inequality and write the solution set in interval notation. Our first step will be figuring out what value of X will make the numerator equal to zero and what value of X will make the denominator equal to zero. We need to first move positive three. From the right hand side to the left hand side of our inequality, we can now multiply negative three here by a factor of X Plus two over X Plus two to achieve a common denominator. Now that we have our common denominator, we can combine our two fractions and simplify the numerator. It's important to note that we have a negative sign out front of this fraction here. Additionally, we have a factor of three multiplying the quantity X plus two. So when we simplify for our numerator, it's important to know that we're going to distribute negative 3/2 X and two positive too. So we're gonna continue algebra over here, where we have one minus three x minus six over X plus two and we can simplify this numerator even further. So we end up with negative three X minus five all over X plus two is greater than or equal to zero. So now that we have a simplified inequality here, we can figure out what values of X will make the numerator and denominator equal to zero. We can recognize that in the numerator here, when we use an X value of negative five thirds, the numerator become zero, and the same is true for a denominator. When X is equal to negative two, these two values are called endpoints, and they helped to break up our solution set into intervals. We can now construct a table using these intervals and employ test points to determine which intervals fall into our solutions. That so he recover table with intervals. They're dependent on our UN points that we determined up above. We contest a value within each interval that is not equal to either end point of the interval in order to see if the test point makes our inequality true or false. If the inequality is true within a certain interval, that means that that interval is included in our solution set. We can start with her first test point of negative three and plug into our simplified inequality the re pluck in in the numerator. We get positive for Emery plug in in the denominator, we get negative one. So when we simplify, we get that negative four is greater than or equal to zero. And this is false. We can now test our second test point of negative 1.8. When we plug into our numerator, we get positive 0.4 and in the denominator, we get positive point to. So when we simplify here, we get that positive two is greater than or equal to zero. And this is true. We can repeat this procedure with our last test point of zero. So from our three test points, we can see that this honorable here is the only interval included in our solution set. So we can start to write our solution and interval notation. And in terms of our brackets here, we can recognize that we haven't equal to component and are inequality. So that usually tells us that our endpoints are included in our solution set. We make an exception when we have an end point that comes from our denominator. And this is because if we were to plug and negative two for X here. In our denominator, it would result in zero, so we would end up dividing whatever our numerator is by zero. And that would give us something undefined on the left hand side of her inequality. And since it's undefined, we can't include the point of negative to in our solution set. So we'll use the heart bracket for negative five thirds and a soft bracket for negative too. And this is our final solution in interval notation.

Okay, so we have to solve the falling inequality. So let's multiply out and they get a three and two. So we get negative six X and then plus 15. That's greater than two plus two and in minus seven. So that's actually minus five. OK, and now let's, um I'm gonna add six x so we can isolate our X on one side. And then I'll add this five on the other side. So yet 20 is greater than a text that is X. Actually, we have been we need to divide by eight. Sort of 20/8 is greater than X or X. Must be less than, um 20. Over eight. 20 is five times for eight is four times two, so we get X must be less than five over for


Similar Solved Questions

5 answers
Reagents and properties Substancequantitymolar mass (g/mol)density (glL)Carboxylic acids Acetic acid, glacial Propionic acid Butyric acid Anthranilic acid Benzoic acid Salicylic acid1.049 0.993 0.964 solid solid solid
Reagents and properties Substance quantity molar mass (g/mol) density (glL) Carboxylic acids Acetic acid, glacial Propionic acid Butyric acid Anthranilic acid Benzoic acid Salicylic acid 1.049 0.993 0.964 solid solid solid...
5 answers
Question4KrusDuring which stage of meiosis would You expect to see nuclear membranel breakdown and crossing over between nonsister chromatids of homologous chromosomes?Zygotene substage of prophaseDiakinesis substage of prophaseLeptotene substage of prophasePachytene substage of prophaseQuestion 54es
Question4 Krus During which stage of meiosis would You expect to see nuclear membranel breakdown and crossing over between nonsister chromatids of homologous chromosomes? Zygotene substage of prophase Diakinesis substage of prophase Leptotene substage of prophase Pachytene substage of prophase Quest...
5 answers
4. Definite integral 1, = f &x (r+a)Vr-
4. Definite integral 1, = f &x (r+a)Vr-...
4 answers
T + 2y + 22 = 1 I + ay + 32 = 3 I + Ily + a2 = bWrite the system in the matrix form AX = B. Perform TOW operations 0n the augmented matrix AIB to transform into the row echelon form_For which values of a does the system have unique solution? for which pairs of values (a,b) does the system have more than one solution? The value of b does not have any effect on whether the system has a unique solution_ Why?
T + 2y + 22 = 1 I + ay + 32 = 3 I + Ily + a2 = b Write the system in the matrix form AX = B. Perform TOW operations 0n the augmented matrix AIB to transform into the row echelon form_ For which values of a does the system have unique solution? for which pairs of values (a,b) does the system have mor...
5 answers
Fnd te eratt Iocation d all the roltive Fande atsolule Axtrema the funatlon in}-Y 4+ with domnainanswens (rom smallyarocsfhas C3uedJakm)fhas (Sala(has [z
Fnd te eratt Iocation d all the roltive Fande atsolule Axtrema the funatlon in}-Y 4+ with domnain answens (rom smally arocs fhas C3ued Jakm) fhas (Sala (has [z...
5 answers
In a sample of 244 men, 73 had high cholesterol levels (more than 200 milligrams per deciliter) In a sample of 232 women, 44 had high cholesterol levels. You want to test the hypothesis that the proportion of people with high cholesterol levels is greater for men than women in the population: State the Null and the Alternate Hypotheses.Ho: P1 pz Hj: P1 P2Ho: P1 = P2 Hi: P1 PzHo: P1 P2 Hi: P1 P2Ho: P1 P2 Hi: P1 = P2
In a sample of 244 men, 73 had high cholesterol levels (more than 200 milligrams per deciliter) In a sample of 232 women, 44 had high cholesterol levels. You want to test the hypothesis that the proportion of people with high cholesterol levels is greater for men than women in the population: State ...
5 answers
Evaluate the integral. (Use 'for the constant of integration:)8 tan € sec? € dx
Evaluate the integral. (Use 'for the constant of integration:) 8 tan € sec? € dx...
5 answers
X+z JS- HW < 2-Xye Ed 97 21 7
X+z JS- HW < 2-Xye Ed 97 21 7...
5 answers
Conducting spherical shell with inner radius and outer radius b concentric with _ larger conducting spherical shall Kaun ua tnz and outer radius such that a b=3cm; € = cm andd = The inner shell positively charged with _ total chargc of +1 HC and the outer shell also positvely charged wth tctl danua 0i *42What is the total charge on the outer surface Dfthe Varge shellVc?
conducting spherical shell with inner radius and outer radius b concentric with _ larger conducting spherical shall Kaun ua tnz and outer radius such that a b=3cm; € = cm andd = The inner shell positively charged with _ total chargc of +1 HC and the outer shell also positvely charged wth tctl ...
5 answers
8 (12 points) If f is entire and |f(2)l < VF , prove that f(2) = 0 for all 2 € C.
8 (12 points) If f is entire and |f(2)l < VF , prove that f(2) = 0 for all 2 € C....
5 answers
You flip a biased coin 30 times The coin will come up heads 70% of the time: What is the mean number of heads you will get?(Binomial Distribution)300.300.7021
You flip a biased coin 30 times The coin will come up heads 70% of the time: What is the mean number of heads you will get?(Binomial Distribution) 30 0.30 0.70 21...
5 answers
Chaptcprobicmcnaikboaro VidcoYour answierpartialk correct Try agalnSunlight, #hose visible wavelengths range (rom 380 750 nmt incicent ne tutace Eheondese #avelenotn (nal WI CAVSE phcroelectronacodium {urtace The work function tor sodium Hemred2.28 eV: Find (4} the maximum kinetic encrgy KEmarThe photoelectton? emiteed(2) NumbeUniNo MnitsTb) NumbeUnits5439Click You would Iiko to Show Work for this qucstion: Opcn ShoxWog
Chaptc probicm cnaikboaro Vidco Your answier partialk correct Try agaln Sunlight, #hose visible wavelengths range (rom 380 750 nmt incicent ne tutace Eheondese #avelenotn (nal WI CAVSE phcroelectrona codium {urtace The work function tor sodium Hemred 2.28 eV: Find (4} the maximum kinetic encrgy KEma...
5 answers
14_ Find 90% confidence interval of the population (true) mean pof warpwise breaking strength if n = 100. A. (178.0675, 183.9034) B. (179.1775, 180.8225) C. (156.0876, 202.9974) D. (13.2987 67.6692)
14_ Find 90% confidence interval of the population (true) mean pof warpwise breaking strength if n = 100. A. (178.0675, 183.9034) B. (179.1775, 180.8225) C. (156.0876, 202.9974) D. (13.2987 67.6692)...

-- 0.023830--