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solve the inequality. graph your solution. |x+3| > or equal to 15

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Hint: |x| > k breaks into x < -k or x > k where k is some positive number
is it still correct if the arrows go in the same direction when I graph it?
no the arrows should be going in opposite directions

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Other answers:

oh so you flip the sign when you make k a negative?
what did you get
x> or equal to 12 and x < or equal to -18
good, you got it
thanks :D
you're welcome
find the slope of the line passing through the given points. then tell whether the line rises, falls, is horizontal, or is vertical. (-7,4) (5,-3)
use the slope formula m = (y2 - y1)/(x2 - x1)
in this case (x1,y1) = (-7,4) (x2,y2) = (5,-3)
how do I find out whether the line rises, falls..etc ?
what did you get for the slope
oh so the line decreases bc it has a negative slope?
you nailed it
Do you mind if i ask you questions from time to time ? I'm trying to study for a trig final on monday.
sure I can help
thanks! alright so this question has the same guidelines as the one we just did but with these coordinates (-5/4, 3); (2/3, 3) my friend had an error in her calculator trying to find the slope , but I got m=0 confused
you're correct, it's 0 how do I know so quickly? notice that the y coordinates are both equal to 3 since the y coordinates are equal, the slope will always be 0 because you have a horizontal line through y = 3
all horizontal lines have a slope of 0
using m = (y2 - y1)/(x2 - x1) we get m = (y2 - y1)/(x2 - x1) m = (3 - 3)/(2/3 - (-5/4)) m = 0/(2/3 - (-5/4)) m = 0
oh ok i understand
what if the x coordinates are equal? would that make the line vertical?
and the slope would be undefined
oh okay
how would i graph y=|x-3|?
do you know how to graph y = | x |
i think so? would there be 2 lines?
yes here is your blank xy axis |dw:1359244103163:dw|
the graph of y = | x | looks like this |dw:1359244135745:dw|
to graph y=|x-3|, you just shift over y = |x| three units to the right
im not sure i get you there...
imagine you can pick up that v shaped graph and move it around well just move it over 3 to the right units like this |dw:1359244315265:dw|
but how do i know the coordinates to create the v shape
i understand the moving it to the right part
the sharp point on the V of y = |x| is at (0,0) move that point over 3 units to the right to get (3,0)
yes i understand that but what are the other coordinates for that shape?
for (3,0)..
(1,1) is on the original graph y = |x| move that 3 units to the right to get (4,1)
(-1,1) is on the original graph y = |x| move that 3 units to the right to get (2,1)
ohhhh so the slope for it is rise 1 over 1?
for the right half of it, yes
left half, slope is -1 |dw:1359244745927:dw|
ohhh ok so on the left side for any problem i will always decrease 1 and rise 1?
im trying to understand the concept so i can graph any absolute value problem
on the left side, you go down 1 and over to the right 1 on the right side, you go up 1 and over to the right 1
the problem they gave me was y=|x-3| would you have to move to graph from (0,0) to (0,-3) ? you moved it up 3
if it was |x| - 3, then you would move (0,0) to (0,-3)
but it's actually |x-3|
would i be able to just plug this in to my calculator ?
yes you should be able to
do you know how to graph absolute value equations?
I guess I forgot :/
what kind of calculator do you have
i just found the absolute value button on my TI-83
and i put it into the y= and i got what you showed me earlier
ok great, glad you did
Solve the equation. \[4x ^{2}-6x+9=0\]
you can either factor or use the quadratic formula
oh wait, one sec
no you can't factor, so you'll have to use the quadratic formula
how do you know you cant factor this problem?
D = b^2 - 4ac D = (-6)^2 - 4(4)(9) D = 36 - 4(4)(9) D = 36 - 144 D = -108 because the discriminant is negative, there are no real solutions so that's why you can't factor
oh okay, and i can use the discriminant as a test on any problem that looks like this one?
yes if the discriminant is a) positive or zero and b) a perfect square then the quadratic can be factored
ohhhhh okay
solve the system of linear equations using any method. 2x+3y=47 7x-8y=-2 2x-y+3z=-19
solve the first two equations for x and y once you have x and y, use them to find z
im not sure how i ended up with 111 for y... the answers say its wrong
no y should be 9, try it again
i did elimination with the first two equations
to get them to equal to 14 and -14 so they can cancel out but i keep getting 111
2x+3y=47 7x-8y=-2 8(2x+3y)=8*47 3(7x-8y)=3(-2) 16x+24y=376 21x-24y=-6 ------------- 72x+0y = 370 37x = 370 x = 370/37 x = 10 2x+3y=47 2(10)+3y=47 20+3y = 47 3y = 47-20 3y = 27 y = 27/3 y = 9 So x = 10 and y = 9 Use them to find z
Oh! I understand now, thank you so much :)
you're welcome

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