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1.Solve the system of equations.

y = 2x^2 - 3
y = 3x - 1
a. no solution
b. (-1/2, 5), (2, -5/2)
c. (-1/2, -5/2), (2,5)
d. (1/2, 5/2), (2, 5)
2.how many real number solutions does the equation have 0 = -3x^2 + x - 4
a. 0
b. 1
c. 2
d. 3
3. solve the equation by completing the square. If necessary round to the nearest hundredth.
x^2 - 18x = 19
a. 1; 19
b. -1; 19
c. 3; 6
d. -3; 1
4. solve. x^2 - 81 = 0
a. 0
b. -9
c. -9, 9
d. 9
5. which model is most appropriate for the data shown in the graph below? (need wedsite to know the problem)
a. quadratic
b. linear
c. exponential
d. line
for questions 6-9, match the equation to its corresponding graph.
(need wedsite to know the problem)
a, b, c, d
6. y=-2x^2+2
7. y=-x^2
8. y=2x^2
9. y=3x^2-4
Graph the quadratic functions y = -2x^2 and y = -2x^2 + 4 on a separate piece of paper. Using those graphs, compare and contrast the shape and position of the graphs.
(text box under question)

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8 answers

  1. Here are the correct answers;
    1. C (-1/2, -5/2), (2,5)
    2. A 0
    3. B -1;19
    4. C -9,9
    5. C Exponential
    MATCHING
    6. A
    7. D
    8. C
    9. B
    10. ESSAY

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  2. i know this is cheating but i need the answers please

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  3. For the essay question
    10.Graph the quadratic functions y=-2x^2 and y=-2x^2+4 on a separate piece of paper. Using those graphs, compare and contrast the shape and position of the graphs.
    Answer: The first function has a vortex at (0,0), while the second function has a vortex of (0,4). The difference between the two is that the second one is higher up on the y-axis than the first one, that it intercepts at a higher point than the first. They are the same without that difference. They both have the same shape too.

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  4. y=3x+2
    4x+2y=24

    helppp

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  5. RO is 100% right. thanks RO :)

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  6. L Writeacher

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  7. I will not give you the answers, but tell you what you have to do

    1. y = 2x^2 - 3
    y = 3x - 1
    both right sides are equal to y, so they must be equal to each other
    2x^2 - 3 = 3x - 1
    2x^2 - 3x - 2 = 0
    this factors nicely

    2.
    0 = -3x^2 + x - 4
    3x^2 - x + 4 = 0 , a=3, b= -1, c = 4
    b^2 - 4ac = 1 - 4(3)(4) = -47
    you would be dealing with √-47 which is not a real number, so what do you think?

    3. x^2 - 18x = 19
    take half of the -18, square that, and add it to both the left and the right side.
    x^2 - 18x + 81 = 19 + 81
    (x-9)^2 = 100
    carry on

    4. x^2 - 81 = 0
    difference of squares ...
    (x+9)(x-9) = 0

    and the solutions are .....

    5. ...... ?

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    Reiny
  8. Absolutely not!

    But if YOU post what YOU THINK, someone might help you.

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    Writeacher

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