Recent content by hackedagainanda

  1. hackedagainanda

    Partial Fraction Decomposition

    I think I got it Ax + Bx = 0 and adding -Ac + Bc =1 A-A =0 add the B's and get 2B = 1 so B= 1/2 and A = -1/2, there is no x term in the numerator so we can move along to the variable c, -Ac = -1/2c and B = 1/2c so that lines up with the books answer. Thanks for the suggestion and help, did I...
  2. hackedagainanda

    Partial Fraction Decomposition

    ##\frac {1} {x^2 -c^2}## with ##c \neq {0}## So the first thing I do is split the ##x^2 -c^2## into the difference of squares so ##x +c## and ##x - c## I then do ##\frac {A} {x + c}## ##+## ##\frac {B} {x-c}##, and then let ##x=c## to zero out the expression. And that is where I am getting...
  3. hackedagainanda

    Intro Math Introductory Probability without Calculus

    Are there any good introductory texts aimed at students with only a working knowledge of High School Algebra? I currently have: Probability: An Introduction by Samuel Goldberg I do plan on eventually learning Calculus, but I would like to start learning probability sooner.
  4. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    (1/2)x(1.5x) So I get 2(1/2)x(1.5x) so adding the area of the rectangle I get 3x^2 + 6x + 1.5x for a total of 4.5x^2 +6x - 336 = 0. The quadratic formula gets me 8, and - 28/3. The area of each triangle is (1/2)8 x 12 which is 48. The area of each triangle is 48 ft^2 both combined is 96 ft^2...
  5. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    Is the quadratic 6x^2 + 6x - 336 = 0 the wrong equation?
  6. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    The 189 ft^2 is correct though... Which is the correct answer, but did I get the right answer for the wrong reason?
  7. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    It does check 3x 7 x (7 +2) = 21 x 9 = 189, + 3(7^2)= 147. 147 + 189 = 336. So the area of the rectangle is 189 ft^2.Thanks for the help, I'll have to remember draw diagrams on the test.
  8. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    The dimensions are 3x(x +2) and the truss is 2 triangles so its x(3x) so its 3x^2 + 3x^2 +6x = 336 Write in standard form and get 6x^2 +6x - 336 = 0 The quadratic formula gets me x = 7, and x = -8
  9. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    The other questions in this section usually say round to the nearest tenth, hundreths, etc. when its a non-integer. Can I get a hint at formulating the problem?
  10. hackedagainanda

    Maximizing Area: Solving Quadratic Applications with Triangles and Rectangles

    So the top of the structure is a triangle with height x. and the height of the rectangle is 2 + x, and the length is 3x. I'm unsure where to go from here. I tried using the formula and getting 3x^2 + 6x +x = 3x^2 + 7x -336 =0 I applied the quadratic formula but it gave me non-integer solutions...
  11. hackedagainanda

    Comparing Solutions of Quadratic Equations: Real vs Imaginary Roots

    Thanks for the help! You are very appreciated :smile:
  12. hackedagainanda

    Comparing Solutions of Quadratic Equations: Real vs Imaginary Roots

    That's the answer I got, I take it the book rationalized the denominator. I see my error now, I didn't follow the steps all the way through.
  13. hackedagainanda

    Comparing Solutions of Quadratic Equations: Real vs Imaginary Roots

    I subtract 5 from both sides to get 7x^2 = -5 Then I divide both sides by 7 to get -5/7. I then take the square root to get x = sqrt of the imaginary unit i 5/7 then ##\pm { i \sqrt \frac 5 7}## The quadratic formula on the other hand gets me a different answer, the discriminant = -140 which...
Back
Top