Posted by Krystle on Friday, April 4, 2008 at 8:10am.
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)^24
2. Find the vertex & axis of symmetry: y=x^210x+3
3. Simplify: (2i+5)(3+4i)
4. Simplify: 3(21+2i)+(45i)
5. Simplify: (27x^2y^3)^3/4
6. Simplify: 5ã300x^42ã243x^4
7. Simplify: 3ã3(4ã3+5ã2)
8. Simplify: (2ã3+ã6)^2
9. Simplify: ^4ã50 divided by ^4ã2
10. 7 divided by ^3ã4
11. 4 divided by ã52

Intermediate Algebra  Krystle, Friday, April 4, 2008 at 8:16am
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.

Intermediate Algebra  marge, Friday, April 4, 2008 at 8:51am
To 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 9:46am
the first two questions form the basics of this topic on quadratic equations.
If you cannot determine the vertex from
y = 5(x+4)^24
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^42(√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, don't know what you mean by
^4√50, there is no base.
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