basic algebra 1 smaple test - 9 Express as a single power...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
22M:001 Basic Algebra I - Final Exam 1. Find an equation for the line passing through the point (1, 2) and perpendicular to the line with equation: 2x + 3y = 1 2. Solve for B: A = (B + b)h. 3. Solve the system: x + y = 3 and -3x + y = 1 4. Subtract and leave in lowest terms: . 5. Solve the equation 2x 2 + 6x = 10. 6. The longer leg of a right triangle is two meters longer than twice the short leg. The hypotenuse is two meters less than three times the short leg. Find the lengths of all three sides of the triangle. 7. Factor as completely as possible: 8x 2 y – 14xy 3 + 6y 5 . 8. Simplify by writing as a rational expression and canceling common factors:
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 9. Express as a single power of m the expression . 10. Solve the equation (if possible) = . 11. Multiply and combine like terms: (x 3 + 1)(x 6 – x 3 + 1). 12. Solve 2x 2 + 6x < -4 and graph the solution set. 13. For what values of k does the equation 2x 2 + 3x + k = 0 have no real solution? 14. A 60% solution of salt is to be mixed with an 80% solution of salt to get 40 liters of a 65% solution of salt. How many liters of each solution should be used? 15. Graph y – 1 = 2x 2 – 4x....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern