anonymous
  • anonymous
solve the inequality. graph your solution. |x+3| > or equal to 15
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
Hint: |x| > k breaks into x < -k or x > k where k is some positive number
anonymous
  • anonymous
is it still correct if the arrows go in the same direction when I graph it?
jim_thompson5910
  • jim_thompson5910
no the arrows should be going in opposite directions

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

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