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thanks i guess i was using the very very long way
how many different ways can the numbers 1,2,3,4 be arranged if you use each number only once?
i got
1,2,3,4
4,1,2,3
3,4,1,2
2,3,4,1
2,1,4,3
1,4,2,3
3,1,4,2
4,3,2,1
any other ways?
and how many can there be....if we just guessed?
a = 3
so put in the value in the equation:
3a - 4c = 7
3(3) - 4c = 7
9 - 4c = 7
subtract 9 both sides
-4c = -2
divide
c = 0.5 or 1/2 right
then c = 0.5
put in value
b - 2c = 1
b - 2(0.5) = 1
b - 1 = 1
add 1 both sides
b = 2 correct
THANKS
it cant be this way right?
a - b = 1
- b - 2c = 1 cross out the b's
------------------
a - 2c = 0 ????
ok i know how to solve system of equations but with this one i am having problems !
1) a - b = 1
2) b - 2c = 1
3) 3a - 4c = 7
ok i know that
you have to first
1equation subtract 2equation
and then
2equation plus 3equation
i know that
but you see that in each equation one letter is missing
so how can we
subtract a-b=1 from b-2c=1
is we dont have a or c????
and same goes for the second part
help:/
ok i understand now thanks you guys
i dont get it
ok so the inverse is?
if
2/2 = 1
so
x = y - 1
then
y = x - 1 as inverse ??
find the inverse function of
y = 2x + 2
original function
y = 2x + 2
rearranged
2x = y - 2
solved for x
x = 1/2y - 1/2
inverse function changed x to y and y to x
y = 1/2x - 1/2
correct
yes no?
take the graph of the function
f(x) = (x - 2)^2,
translate it two units to the right, and write the equation of the new graph.
how do i do that?
they gave me two graphs:/
use the quadratic formula to find the roots of the equation
3x^3 - 2x + 5 = 0
i did
formula :
-b +- sqrt (b^2 - 4ac)
x = ----------------------------
2a
- (-2) +- sqrt( (-2)^2 - 4(3)(5) )
x = ----------------------------------------
2(3)
2 +- sqrt (-4 - 4(15) )
x = ------------------------------
6
2 +- sqrt ( -4 - 60)
x = ------------------------------
6
2 +- sqrt ( -64 )
x = ---------------------------
6
now what i do i forgot?
oh i see
i never heard of those before.
hm
lol nice
crackers?
oh really.
yeah i have been using imageshack very much..to upload graphs and problems such as that. its very useful
cool
thanks
ok to post any image...
i coudlnt copy from my mac to here
so i used this site to upload my image and copy the direct link image ...and tehn i put img the link and img again
and it came up
here is the site
http://www.imageshack.us/
OHHHHHHHH MY GODDDDDDDD.it worked
ok let me try..right now to post one pic
yeah maybe your right
unique wrote:...but i took so great pictures..a great photographer if i do say myself.
Please share them. When you click "Post Reply" there is an image upload where you can do several at once.
umm ok i'll be happy to share them....
if the img share tab works for me