please help......
((2z-8)/(z^2-4))/((z^2+6z+8)/(z-4))
simplify completly..... show work

- anonymous

please help......
((2z-8)/(z^2-4))/((z^2+6z+8)/(z-4))
simplify completly..... show work

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- anonymous

do you know how to do this?

- dumbcow

First Flip bottom fraction and Multiply
Then factor and things will cancel
\[2z-8 = 2(z-4)\]
\[z^{2}-4 = (z+2)(z-2)\]
\[z^{2}+6z+8=(z+4)(z+2)\]

- anonymous

wow i forgot that the things cancelled.... okay what did you get for your answer?

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## More answers

- anonymous

im not sure how you did that can you show me?

- dumbcow

\[=\frac{2(z-4)(z-4)}{(z+2)(z+2)(z-2)(z+4)}\]
hmm maybe they dont cancel in this case..haha

- anonymous

yea and i think you factored wrong.....

- anonymous

wait how did you factor that?

- dumbcow

no everything is factored correctly

- dumbcow

which one

- anonymous

how did you do it then? i got something completly different but i thrust what you got....

- dumbcow

are you sure its division and not multiplying the 2 fractions ??

- anonymous

you flip the second one and change to multi.

- dumbcow

correct

- anonymous

and then can you show me what you did im completly lost

- dumbcow

oh for the factoring part.
well for 2z -8 i just pulled out a 2 from each term
2(z-4) = 2z - 8

- anonymous

ok

- dumbcow

For z^2 - 4,
find factors of -4 that add up to 0 since there is no "z" term
-2*2 = -4 and -2+2=0
(z-2)(z+2)

- anonymous

wow thats sooo much easier now thanks :)
can you help me with a couple more?

- dumbcow

z^2 +6z +8
same thing, look for factors of 8 that add up to 6
4*2 =8 and 4+2 =6
(z+4)(z+2)

- dumbcow

ok

- anonymous

okay.......
((4)/(2x+1))-((3)/(2x))=

- anonymous

can you show me step by step please?

- dumbcow

adding/subtracting fractions you need to get common denominator
2x+1 cannot be factored
2x cannot be factored
common denominator is 2x(2x+1)
just like adding 1/2 + 1/3, common denominator is 2*3=6

- dumbcow

now change numerators
\[\frac{4}{2x+1} = \frac{(2x)(4)}{2x(2x+1)}\]
\[\frac{3}{2x} = \frac{3(2x+1)}{2x(2x+1)}\]
combine into 1 fraction
\[\frac{(2x)(4) - 3(2x+1)}{2x(2x+1)}\]

- anonymous

ok

- dumbcow

multiply and add like terms on top
\[\frac{8x -6x -3}{2x(2x+1)} = \frac{2x-3}{2x(2x+1)}\]

- anonymous

can you help me with this one?
((m+5)/(2m^2-2))+((3)/(1-m))+((5)/(2m+2))

- dumbcow

factor all the denominators
2m^2 -2 = 2(m-1)(m+1)
1-m = -(m-1)
2m+2 = 2(m+1)
common denominator will include everything w/out repeats
-2(m-1)(m+1)

- anonymous

okay i can do it from there can you help me with this one?
((n+1)-(2/n))/((n+4)+(4/n))

- dumbcow

\[=\frac{-(m+5)+(2(m+1))(3) + (-(m-1))(5)}{-2(m+1)(m-1)}\]

- anonymous

can you help me with this one? please????
((n+1)-(2/n))/((n+4)+(4/n))

- dumbcow

ok combine fractions on top and bottom, then flip and multiply

- dumbcow

common denominator will just be n

- anonymous

ok

- dumbcow

answer should be (n-1)/(n+2)

- anonymous

yup that what i got are you good with summation notation and geometric means/sequences?

- dumbcow

sure

- anonymous

okay.... wellthe question is...
insert four geometirc means between -7 and -224

- dumbcow

hmm dont understand?

- anonymous

nevermind i figured that one out......

- anonymous

what about....

- anonymous

\[\sum_{n=1}^{50}(1/4)(n+2)\]

- dumbcow

\[=\frac{1}{4}\sum_{n=1}^{50}n+2 = \frac{1}{4}(\sum_{n=1}^{50}n +\sum_{n=1}^{50}2)\]

- anonymous

im stuck on this one..
which term of the geometric sequence 243, -81, 27, . . . is (-1/9)?

- dumbcow

8 th term

- anonymous

how did you figure that out?

- dumbcow

haha tried proving it using geometric sequence formula but got stuck when taking log
anyway just continue the sequence of dividing by 3 and flipping the sign

- anonymous

o so you flip the sign every other time?
could you help me with this one?
there are seven houses; in each are seven cats. each cat kills seven mice. each mouse would hvae eaten seven ears of wheat. each ear of wheat prodece seven measures of grain. how much grain is saved?
i got 16807 grains saved

- dumbcow

correct
7^3 mice
7^4 ears
7^5 grain

- anonymous

ok that equals 16807 which is what i got:)

- anonymous

okay change the repeating decimal .25 (sopposto have the bar above the 25) to an equivlant common fraction

- dumbcow

?? why do you need to know that
some fraction close to proportional of 1/4

- anonymous

i dontk know its on my homework though ydo you know what it'd be?

- dumbcow

haha i just guessed and put it in my calculator
25/99

- anonymous

wow your good.... do you know how i would show that? just say guess and check? we turn this in for a grade thats why im asking

- dumbcow

actually any n/99 = .nnnnnn

- anonymous

in a certain credit union, money left on deposit for one year earns 4% intrest at the end ot the year. if you invested $100 at the beginning of each year in this credit union and did NOT withdraw the intrest due at the end of the year, how much would you hvae on deposit at the end of the tenth year?

- dumbcow

=100(1.04 + 1.04^2+... +1.04^10)
need sum of geometric sequence
sum = a(1-r^n)/(1-r)
sum = 1.04(1-1.04^10)/(1-1.04)

- dumbcow

i get 1248.64

- anonymous

so when you get your answer are you sopposto multiply it by 100?

- dumbcow

yes

- anonymous

okay just stay right there i have a couple more i just have to go and get another pencil mine just broke

- anonymous

ok im back i dont know what formula you used

- dumbcow

for finding sum of geometric sequence

- anonymous

yea but can you tell me it with all of the variable in it please? i need to look and see if i have that one

- dumbcow

\[s _{n} = \frac{a _{1}(1-r ^{n})}{1-r}\]

- anonymous

i only have
SofN=(Asub1-Asub1R^n)/(1-R)
and...
SofN=(Asub1-AsubnR)/(1-R)
sooo.. which one should i use, and what numbers?

- dumbcow

ues the first one, only difference is mine has factored out the a1 on top

- dumbcow

a1 = first term
r = common ratio

- anonymous

ummmm... i should get the same answer right?
i think im doing someting wrong im getting 1200.610712

- dumbcow

1-(1.04)^10 = -.4802
1.04(-.4802) = -.49945
-.49945/(1-1.04) = 12.48635

- anonymous

wait shouldnt it be.....
(100-100(1.04)^10)/(1-1.04)

- dumbcow

no im leaving the 100 on outside of sequence
then we multiply the sum by 100

- anonymous

but when i do it shouldnt i be getting the same thing? and i dont understand how your doing it

- dumbcow

ok if you include the 100 then sequence will look like this
=100(1.04) + 100(1.04^2) +...+100(1.04^10)
a1 = 100(1.04)
r = 1.04
sn = 100(1.04)(1-1.04^10)/(1-1.04)

- anonymous

but my equation is Sn=(A1-A1R^n)/(1-R)

- dumbcow

same thing
factor out the A1

- anonymous

can you do it my way when you factor out the A1 it confuses me

- anonymous

....^because when you factor....

- dumbcow

ok
Sn = 100(1.04) -100(1.04)(1.04^10) / (1-1.04)

- anonymous

ok ...
a ball which rolls off a penthouse terrace falls 16 feet the first second, 48 feet the next second, and 80 feet the third second. if it continues to fall in this mannor, how far does it fall in the seventh second?

- dumbcow

i believe this models a parabola of y=-16x^2
the change in y from 6 to 7 is how far it falls in 7th second
16(7^2) - 16(6^2) = 16(49-36) = 16(13) = 208

- anonymous

a rubber ball dropped 40 feet rebounds on each bounce 2/5 of the distance from which it fell. how far will it travel before comming to a rest?

- dumbcow

geometric sequence
40+40(2/5)+ 40(2/5)^2 +...
looks like an infinite sequence before you get 0
lim n->inf (2/5)^n = 0
so in formula r^n = 0
a1=40
r=2/5
sum = 40 - 40(0) / 1-(2/5)
sum = 40 /3/5 = 200/3 = 66.666

- anonymous

what formula is that?

- dumbcow

same as before
sum of geometric sequence

- anonymous

no but which one of mine because yours confuse me

- dumbcow

\[s _{n}=\frac{a _{1}-a _{1}r^{n}}{1-r}\]

- anonymous

what would i put for n?

- dumbcow

infinity

- anonymous

hu?

- dumbcow

just substitute r^n=0

- anonymous

would my answer be iin feet?

- dumbcow

yes

- anonymous

what does the graph for ..... look like?
-cubic with one real solution and two complex solutions look like?
-cubic with no real solutions?
-quartic with no real solutions?
-a quadratic with one real solution and one complex solution?
-a quartic with two real solutions and two complex solutions?
-a quadratic with no real solutions?
-a quartic with no real solutions?
-a cubic with 3 real solutions, but one is a double root?
-a quartic with for real roots, but both are double roots?

- dumbcow

umm number of real solutions represents number of x_intercepts
cubic with no real solutions does not exist i believe

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