Search
- Filter Results
- Location
- Classification
- Include attachments
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/04%3A_Quadratic_and_Polynomial_Functions/4.02%3A_Graphs_of_Polynomials/4.2E%3A_ExercisesThe polynomial function h(t)=−16t2+300 gives the height of a ball t seconds after it is dropped from a 100-foot tall bridge. As x→∞, f(x)→−∞ and as x→−∞, \(f(x)\...The polynomial function h(t)=−16t2+300 gives the height of a ball t seconds after it is dropped from a 100-foot tall bridge. As x→∞, f(x)→−∞ and as x→−∞, f(x)→∞ As x→∞, f(x)→∞ and as x→−∞, f(x)→−∞ As x→∞, f(x)→−∞ and as x→−∞, f(x)→−∞ As x→∞, f(x)→∞ and as x→−∞, f(x)→∞
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/00%3A_Front_Matter/05%3A_About_the_AuthorsTeaching intermediate algebra was a lot of fun, so I have been really excited to work on curating this text designed for intermediate algebra students and hope that it will be helpful for any student ...Teaching intermediate algebra was a lot of fun, so I have been really excited to work on curating this text designed for intermediate algebra students and hope that it will be helpful for any student looking to strengthen their skillset, as well as any instructor hoping to assist them in their journey!
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/01%3A_Foundations/1.05%3A_Solve_Linear_Inequalities/1.5E%3A_ExercisesGraph Inequalities In the following exercises, graph each inequality on the number line and write in interval notation. x<−73 −5≤x<−3 0≤x≤3.5 −3.75≤x≤0 \(b...Graph Inequalities In the following exercises, graph each inequality on the number line and write in interval notation. x<−73 −5≤x<−3 0≤x≤3.5 −3.75≤x≤0 b+78≥16 Now with Interval Notation In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. 4k−(k−2)≥7k−26 6n−12(3−n)≤9(n−4)+9n 9u+5(2u−5)≥12(u−1)+7u
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/01%3A_Foundations/1.02%3A_FractionsWhen this happens, it is a good idea to factor the numerator and the denominator into prime numbers using the factor tree, then cancel out the common factors using the Equivalent Fractions Property. T...When this happens, it is a good idea to factor the numerator and the denominator into prime numbers using the factor tree, then cancel out the common factors using the Equivalent Fractions Property. The reciprocal of a fraction is found by inverting the fraction, placing the numerator in the denominator and the denominator in the numerator.
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/05%3A_Everything_else_you_need_to_know
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/05%3A_Everything_else_you_need_to_know/5.04%3A_Multiply_and_Divide_Rational_Expressions\(\boldsymbol{\begin{array} {ll} &\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5} \\ & \\ \begin{array} {ll} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array}} &\dfrac...\(\boldsymbol{\begin{array} {ll} &\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5} \\ & \\ \begin{array} {ll} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array}} &\dfrac{(3a+1)(a−3)(a+5)(a+5)}{(a−5)(a+5)(3a+1)(a−5)} \\ & \\ Simplify by dividing outcommon factors. &\dfrac{\cancel{(3a+1)}(a−3)\cancel{(a+5)}(a+5)}{(a−5)\cancel{(a+5)}\cancel{(3a+1)}(a−5)} \\ & \\ \text{Simplify.} &\dfrac{(a−3)(a+5)}{(a−5)(a−5)} \\ & \\ \…
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/04%3A_Quadratic_and_Polynomial_Functions/4.05%3A_Solve_Quadratic_Equations_Using_the_Quadratic_FormulaWe have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific vari...We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. To use the Quadratic Formula, we substitute the values of a,b, and c from the standard form into the expression on the right side of the formula. The solutions to a quadratic equation of the form ax2+bx+c=0,a≠0 are given by the formula:
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/03%3A_Linear_Equations_and_Graphs/3.02%3A_Graphing_Linear_Equations/3.2E%3A_ExercisesIn the following exercises, graph by plotting points. In the following exercises, graph each equation. In the following exercises, graph each pair of equations in the same rectangular coordinate syste...In the following exercises, graph by plotting points. In the following exercises, graph each equation. In the following exercises, graph each pair of equations in the same rectangular coordinate system. y=2x and y=2 y=−12x and y=−12 In the following exercises, find the x- and y-intercepts on each graph. In the following exercises, find the intercepts for each equation. In the following exercises, graph using the intercepts.
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/05%3A_Everything_else_you_need_to_know/5.06%3A_Solve_Rational_EquationsWe found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions. An extraneous solution to a rational equation is an algebraic ...We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions. An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined. (x+2)(x−2)2x+2+(x+2)(x−2)4x−2=(x+2)(x−2)(x−1x2−4)
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/03%3A_Linear_Equations_and_GraphsIn this chapter, you will explore linear equations, develop a strategy for solving them, and relate them to real-world situations.
- https://math.libretexts.org/Courses/Kansas_State_University/Your_Guide_to_Intermediate_Algebra/04%3A_Quadratic_and_Polynomial_Functions/4.06%3A_Graph_Quadratic_Functions_Using_Properties/4.6E%3A_Exercisesa. f(x)=−2x2−6x−7 b. f(x)=6x2+2x+3 a. f(x)=−3x2+5x−1 b. f(x)=2x2−4x+5 f(x)=x2+10x+25 f(x)=−x2+2x+5 f(x)=x2+7x+6 f(x)=x2+8x+12 \(f(x)...a. f(x)=−2x2−6x−7 b. f(x)=6x2+2x+3 a. f(x)=−3x2+5x−1 b. f(x)=2x2−4x+5 f(x)=x2+10x+25 f(x)=−x2+2x+5 f(x)=x2+7x+6 f(x)=x2+8x+12 f(x)=x2+6x+13 f(x)=4x2−20x+25 f(x)=x2+6x+5 f(x)=x2+4x+3 f(x)=9x2+12x+4 f(x)=2x2−4x+1 f(x)=2x2−4x+2 f(x)=−x2−4x+2 f(x)=5x2−10x+8 f(x)=3x2+18x+20