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April 20, 2014

April 20, 2014

Posted by **Krystle** on Friday, April 4, 2008 at 8:10am.

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: 5300x^4-2243x^4

7. Simplify: 33(43+52)

8. Simplify: (23+6)^2

9. Simplify: ^450 divided by ^42

10. 7 divided by ^34

11. 4 divided by 5-2

- Intermediate Algebra -
**Krystle**, Friday, April 4, 2008 at 8:16amnote: 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.

- Intermediate Algebra -
**marge**, Friday, April 4, 2008 at 8:51amTo answer number 3 use f.o.i.l and then wherever you get isquared turn to a negative 1

(2i+5)(3+4i)

6i + 8i^2 + 15 + 20i

6i +8(-1) + 15 + 20i

6i - 8 + 15 + 20i

26i + 7

- Intermediate Algebra -
**Reiny**, Friday, April 4, 2008 at 9:46amthe 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

or

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, dont know what you mean by

^4√50, there is no base.

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