NatalieandKayla
solve the inequality. graph your solution.
|x+3| > or equal to 15
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jim_thompson5910
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Hint:
|x| > k
breaks into
x < -k or x > k
where k is some positive number
NatalieandKayla
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is it still correct if the arrows go in the same direction when I graph it?
jim_thompson5910
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no the arrows should be going in opposite directions
NatalieandKayla
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oh so you flip the sign when you make k a negative?
jim_thompson5910
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exactly
jim_thompson5910
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what did you get
NatalieandKayla
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x> or equal to 12 and x < or equal to -18
jim_thompson5910
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good, you got it
NatalieandKayla
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thanks :D
jim_thompson5910
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you're welcome
NatalieandKayla
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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)
jim_thompson5910
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use the slope formula
m = (y2 - y1)/(x2 - x1)
jim_thompson5910
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in this case
(x1,y1) = (-7,4)
(x2,y2) = (5,-3)
NatalieandKayla
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how do I find out whether the line rises, falls..etc ?
jim_thompson5910
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what did you get for the slope
NatalieandKayla
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-7/12
NatalieandKayla
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oh so the line decreases bc it has a negative slope?
jim_thompson5910
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you nailed it
NatalieandKayla
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Do you mind if i ask you questions from time to time ? I'm trying to study for a trig final on monday.
jim_thompson5910
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sure I can help
NatalieandKayla
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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 ...im confused
jim_thompson5910
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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
jim_thompson5910
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all horizontal lines have a slope of 0
jim_thompson5910
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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
NatalieandKayla
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oh ok i understand
NatalieandKayla
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what if the x coordinates are equal? would that make the line vertical?
jim_thompson5910
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exactly
jim_thompson5910
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and the slope would be undefined
NatalieandKayla
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oh okay
NatalieandKayla
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how would i graph y=|x-3|?
jim_thompson5910
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do you know how to graph y = | x |
NatalieandKayla
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i think so? would there be 2 lines?
jim_thompson5910
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yes here is your blank xy axis
|dw:1359244103163:dw|
jim_thompson5910
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the graph of y = | x | looks like this
|dw:1359244135745:dw|
jim_thompson5910
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to graph y=|x-3|, you just shift over y = |x| three units to the right
NatalieandKayla
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im not sure i get you there...
jim_thompson5910
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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|
NatalieandKayla
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but how do i know the coordinates to create the v shape
NatalieandKayla
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i understand the moving it to the right part
jim_thompson5910
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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)
jim_thompson5910
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|dw:1359244469161:dw|
NatalieandKayla
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yes i understand that but what are the other coordinates for that shape?
NatalieandKayla
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for (3,0)..
jim_thompson5910
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(1,1) is on the original graph y = |x|
move that 3 units to the right to get (4,1)
jim_thompson5910
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|dw:1359244638800:dw|
jim_thompson5910
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(-1,1) is on the original graph y = |x|
move that 3 units to the right to get (2,1)
jim_thompson5910
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|dw:1359244675721:dw|
NatalieandKayla
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ohhhh so the slope for it is rise 1 over 1?
jim_thompson5910
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for the right half of it, yes
jim_thompson5910
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|dw:1359244723365:dw|
jim_thompson5910
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left half, slope is -1
|dw:1359244745927:dw|
NatalieandKayla
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ohhh ok so on the left side for any problem i will always decrease 1 and rise 1?
NatalieandKayla
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im trying to understand the concept so i can graph any absolute value problem
jim_thompson5910
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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
NatalieandKayla
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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
jim_thompson5910
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if it was |x| - 3, then you would move (0,0) to (0,-3)
jim_thompson5910
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but it's actually |x-3|
NatalieandKayla
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oh
NatalieandKayla
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would i be able to just plug this in to my calculator ?
jim_thompson5910
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yes you should be able to
jim_thompson5910
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do you know how to graph absolute value equations?
NatalieandKayla
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I guess I forgot :/
jim_thompson5910
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what kind of calculator do you have
NatalieandKayla
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i just found the absolute value button on my TI-83
NatalieandKayla
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and i put it into the y= and i got what you showed me earlier
jim_thompson5910
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ok great, glad you did
NatalieandKayla
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Solve the equation.
\[4x ^{2}-6x+9=0\]
jim_thompson5910
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you can either factor or use the quadratic formula
jim_thompson5910
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oh wait, one sec
jim_thompson5910
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no you can't factor, so you'll have to use the quadratic formula
NatalieandKayla
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how do you know you cant factor this problem?
jim_thompson5910
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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
NatalieandKayla
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oh okay, and i can use the discriminant as a test on any problem that looks like this one?
jim_thompson5910
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yes if the discriminant is a) positive or zero and b) a perfect square
then the quadratic can be factored
NatalieandKayla
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ohhhhh okay
NatalieandKayla
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solve the system of linear equations using any method.
2x+3y=47
7x-8y=-2
2x-y+3z=-19
jim_thompson5910
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solve the first two equations for x and y
once you have x and y, use them to find z
NatalieandKayla
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im not sure how i ended up with 111 for y... the answers say its wrong
jim_thompson5910
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no y should be 9, try it again
NatalieandKayla
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i did elimination with the first two equations
NatalieandKayla
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to get them to equal to 14 and -14 so they can cancel out but i keep getting 111
jim_thompson5910
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2x+3y=47
7x-8y=-2
8(2x+3y)=8*47
3(7x-8y)=3(-2)
16x+24y=376
21x-24y=-6
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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
NatalieandKayla
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Oh! I understand now, thank you so much :)
jim_thompson5910
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you're welcome