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