Intermediate Algebra

I have a take home test and I'm stuck on a couple. Anyone help please?

1. Find the vertex & axis of symmetry: -5(x+4)^2-4
2. Find the vertex & axis of symmetry: y=x^2-10x+3
3. Simplify: (2i+5)(3+4i)
4. Simplify: 3(21+2i)+(4-5i)
5. Simplify: (27x^2y^3)^-3/4
6. Simplify: 5�ã300x^4-2�ã243x^4
7. Simplify: 3�ã3(4�ã3+5�ã2)
8. Simplify: (2�ã3+�ã6)^2
9. Simplify: ^4�ã50 divided by ^4�ã2
10. 7 divided by ^3�ã4
11. 4 divided by �ã5-2

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asked by Krystle
  1. note: The a keys are supposed to be square root. I guess the sign doesn't show up here. Also, some are to the power of.

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    posted by Krystle
  2. To answer number 3 use f.o.i.l and then wherever you get isquared turn to a negative 1

    6i + 8i^2 + 15 + 20i
    6i +8(-1) + 15 + 20i
    6i - 8 + 15 + 20i
    26i + 7

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    posted by marge
  3. the first two questions form the basics of this topic on quadratic equations.

    If you cannot determine the vertex from

    y = -5(x+4)^2-4

    I think you are in deep trouble.

    #3 is done for you, do #4 the same way

    #5 (27x^2y^3)^-3/4 = 1/(27x^2y^3)^+3/4
    any more "simplification" would only make it look more complicated

    #6 I cannot tell where your square root ends.
    Is it (5�√300)x^4-2�(√243)x^4
    5�√(300x^4)-2�√(243)x^4) ?

    if the first, then
    =5*(10√3)x^4 - 2(9√3)x^4
    = (32√3)x^4

    #7,8, just expand them

    #9 and #10, don't know what you mean by
    ^4√50, there is no base.

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    posted by Reiny

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