'A. DIRECTIONS: Read each item carefully then writes the letter of the best answer on a sheet of paper DIRECTIONS: Read each item carefully then writes the letter of the best answer on sheet of paper. Which of the following represents quadratic inequality? A.3x2 4x - 5=0B.>_82 3x Cx +4x-5 D. 3x - 5 < 0 2. Which of the following is a solution to the inequality x $-2? A -2 3. How do you write "for all values of x such that x is greater than negative 2 but is less than 6"? A {x:-2sXs6} B: {x: -2 < x < 6} C. {x:-22x2 6} D {x: -2 > > 6} The solution set {x: -3 < x < 1} can be expressed as A. (-3,1] B. [3, 1) C.(-3,1) D.[3, 1] 5 . Find the solution set ofx? 2x - 15 > 0. A {x x <-3 or x > 5} B. {x: -5 < x < 3} C:{xx < =5 or x > 3} D: {x: -< x < 5} Using the given the graphical solution, which statements islare true? I. Zero (0) is one of the solutions set II. Any positive real numbers are the solution set. A. |only B. Il only C.both | and none of the above 7 . Which of the following is the correct interval notation for each region in the number line at the right? 4) U[-t,5]u (5,0) (~o, 4) U (4,5)U (5,0) (~o,-4]u(-4,5) U[5,00) D. (~0,-4]u[-4,5]u [5,0) 8 . Bea needs to make garden plot which has an area less than 24f2 . The length should be 2 ft longer than the width_ What are the possible dimensions of the box? 2ft by 6ft B. 3ft by 4f C.3f by Sft D. 4f by 6 f 9. The product of two integers is at most 50 and their sum is 15. Which of the following is NOT solution? A. (12, 3) B. (11,4) C. (10,5) D. (9,6) 10. The area of rectangular garden should be at least 300 square meters_ If its lengih ( / ) is 10 meters more than twice its width (w) find the range of the possible values for the lots width_ ws10m w < 15m w 2 1Om D. w 2 15m'

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Erika Bustos

The University of Texas at El Paso

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'A. DIRECTIONS: Read each item carefully then writes the letter of the best answer on a sheet of paper DIRECTIONS: Read each item carefully then writes the letter of the best answer on sheet of paper. Which of the following represents quadratic inequality? A.3x2 4x - 5=0B.>_82 3x Cx +4x-5 D. 3x - 5 < 0 2. Which of the following is a solution to the inequality x $-2? A -2 3. How do you write "for all values of x such that x is greater than negative 2 but is less than 6"? A {x:-2sXs6} B: {x: -2 < x < 6} C. {x:-22x2 6} D {x: -2 > > 6} The solution set {x: -3 < x < 1} can be expressed as A. (-3,1] B. [3, 1) C.(-3,1) D.[3, 1] 5 . Find the solution set ofx? 2x - 15 > 0. A {x x <-3 or x > 5} B. {x: -5 < x < 3} C:{xx < =5 or x > 3} D: {x: -< x < 5} Using the given the graphical solution, which statements islare true? I. Zero (0) is one of the solutions set II. Any positive real numbers are the solution set. A. |only B. Il only C.both | and none of the above 7 . Which of the following is the correct interval notation for each region in the number line at the right? 4) U[-t,5]u (5,0) (~o, 4) U (4,5)U (5,0) (~o,-4]u(-4,5) U[5,00) D. (~0,-4]u[-4,5]u [5,0) 8 . Bea needs to make garden plot which has an area less than 24f2 . The length should be 2 ft longer than the width_ What are the possible dimensions of the box? 2ft by 6ft B. 3ft by 4f C.3f by Sft D. 4f by 6 f 9. The product of two integers is at most 50 and their sum is 15. Which of the following is NOT solution? A. (12, 3) B. (11,4) C. (10,5) D. (9,6) 10. The area of rectangular garden should be at least 300 square meters_ If its lengih ( / ) is 10 meters more than twice its width (w) find the range of the possible values for the lots width_ ws10m w < 15m w 2 1Om D. w 2 15m'

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Video Transcript

we have to solve the following. So the first one says quadratic inequality. So it has to have an inequality in the eve the expression So A doesn't see doesn't even have equal or inequality and then for it to be a quadratic it has to be have an X squared so D. Doesn't so letter B is our only choice. So the fall we need to pick which of the following numbers is less than or equal to negative two. Negative two is equal to negative two. Negative one is bigger, sees zero is bigger and one is bigger. And so then we have to write the following and set notation. So we have X is greater than negative two, so greater than negative two but less then six. So that matches answer choice B. Then we have four. There's no equal signs, there's only going to be parentheses in our interval notation. So that throws out a B and D. So we're only left with C. So then we have to factor X squared minus five. X plus factory. So we have X -5, X plus three is greater than zero. So we get X -5. Well, we're going to first think of it as an equation so we can get our solutions. So equal zero. So we get zero and then X plus three is equal to zero. So X is equal to negative five and X is equal to negative three. So we have to have a negative five, -5 or positive five And a -3. So if it doesn't have a positive five and a negative three, it's wrong. So C. Is gone and B is gone. So now we gotta check. So we have our little number line here. So we have negative three and then we have positive five. So if we plug in negative three it makes this equal to zero, It has to be greater than zero. So -3 is not included. And the same thing if we plug in five it's not included. So now we're going to plug in a number that's in the middle to see if we shade less than or greater than. So we're gonna try zero itself. So we plug in zero. We get zero minus 50 plus three. So that's negative five and three. That's -15. Negative 15 is not greater than zero. That's not true. So we can't shade any numbers between -3 and five. So it has to be numbers that are smaller than -3. We gotta shade in this direction and we gotta shade in this direction. So that means that a is answer choice not deep. Then using the given graphical solution, which statements are true. So zero is is one of the solution sets. That's not true because zero is here and the solutions are pointing this way and this way any positive real numbers are in the solution set. Um Any yes any positive number because if we go to this section, that's all positive numbers. So that's of the two statements. That's the only thing that's true. We need to find the correct inevitable notation for number seven. So everything is included except for -4 and five. So it goes from negative infinity to negative four, but negative four is not included. So it should not have a bracket at negative four. So C. Is gone and D. Is gone because they have brackets and then when it starts again it should start with and parentheses not a bracket. So A. Is gone. So that leaves B as our only choice, Then we have number 8B needs to make a garden plot Which has an area less than 20. For the length should be two ft longer than the width. What are the possible dimensions? So to find the area of a rectangle you do length times with So length times width is equal to 24 but it says the length should be two ft longer than the width. So we have with plus two times W is equal to 24 W squared plus two W minus 24. So we have w plus six W minus four. So we have a width of -6 which makes no sense. And we got a with a positive four. So we plug that back in, we get a width of six, six and four and a length of four. So that matches choice D. And that says the product of two imagers is at most 50 and their sum is 15. So when you multiply two numbers it's 50. When you add them it's 15. So we have to solve so we get X. is equal to 15 -Y. So we have X. We're gonna replace sex, we have 15 minus Y Times Y. is equal to 50. So we have negative Y squared Plus 15. Y is equal to 50. We're gonna move everything to the right so zero is equal to 50 square first. Why squared -15? Why plus 50? So this breaks down to y plus minus 10. Why minus five? So we have Y. is equal to 10 and Y is equal to five. It's not a solution. So we have two says the product of two imagers is at most 50 and their sum is 15. So when you multiply Um 12 times three that's 36. um 11 times four. That's 40 for 10 times five Is 50. And then nine times six gives you 54. So that does is not in the solution because that's more than 50. Okay so the area of a rectangle should be at least 30 m2 300 m2. So we have length times width is going to be at least 300. The length is um 10 more than twice the width. So two W plus 10 times w it's going to be greater than or equal to 300. So we have two W squared plus 10 W. We move the 300 over so That's greater than or equal to zero. So we take out a two so we w squared plus five w minus 150. So we have w plus 10 w minus five. So we get one solution to be negative 10 and one solution to B5. So we can't have a negative with. So our with has to be greater than or equal to. So it says the length is twice more than the way to find the range of the possible values of the lot of the that's 10, five. Well that's 50, whoops. I did 50 instead of 1:50. So factors a 1 50 that are going to subtract to give you five. So we have 1 50 at 10 and 15 And left the whole one out. So 10 W plus 15 W minus 10. So this gives you a width of -15, we can't use that one, this gives you a width of 10. So this is the one that we're going to use or with has to be greater than or equal to 10. So that's letter C.

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