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

- anonymous

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

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- jamiebookeater

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- jim_thompson5910

Hint:
|x| > k
breaks into
x < -k or x > k
where k is some positive number

- anonymous

is it still correct if the arrows go in the same direction when I graph it?

- jim_thompson5910

no the arrows should be going in opposite directions

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## More answers

- anonymous

oh so you flip the sign when you make k a negative?

- jim_thompson5910

exactly

- jim_thompson5910

what did you get

- anonymous

x> or equal to 12 and x < or equal to -18

- jim_thompson5910

good, you got it

- anonymous

thanks :D

- jim_thompson5910

you're welcome

- 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

use the slope formula
m = (y2 - y1)/(x2 - x1)

- jim_thompson5910

in this case
(x1,y1) = (-7,4)
(x2,y2) = (5,-3)

- anonymous

how do I find out whether the line rises, falls..etc ?

- jim_thompson5910

what did you get for the slope

- anonymous

-7/12

- anonymous

oh so the line decreases bc it has a negative slope?

- jim_thompson5910

you nailed it

- 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

sure I can help

- 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

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

all horizontal lines have a slope of 0

- 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

oh ok i understand

- anonymous

what if the x coordinates are equal? would that make the line vertical?

- jim_thompson5910

exactly

- jim_thompson5910

and the slope would be undefined

- anonymous

oh okay

- anonymous

how would i graph y=|x-3|?

- jim_thompson5910

do you know how to graph y = | x |

- anonymous

i think so? would there be 2 lines?

- jim_thompson5910

yes here is your blank xy axis
|dw:1359244103163:dw|

- jim_thompson5910

the graph of y = | x | looks like this
|dw:1359244135745:dw|

- jim_thompson5910

to graph y=|x-3|, you just shift over y = |x| three units to the right

- anonymous

im not sure i get you there...

- 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

but how do i know the coordinates to create the v shape

- anonymous

i understand the moving it to the right part

- 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

|dw:1359244469161:dw|

- anonymous

yes i understand that but what are the other coordinates for that shape?

- anonymous

for (3,0)..

- jim_thompson5910

(1,1) is on the original graph y = |x|
move that 3 units to the right to get (4,1)

- jim_thompson5910

|dw:1359244638800:dw|

- jim_thompson5910

(-1,1) is on the original graph y = |x|
move that 3 units to the right to get (2,1)

- jim_thompson5910

|dw:1359244675721:dw|

- anonymous

ohhhh so the slope for it is rise 1 over 1?

- jim_thompson5910

for the right half of it, yes

- jim_thompson5910

|dw:1359244723365:dw|

- jim_thompson5910

left half, slope is -1
|dw:1359244745927:dw|

- anonymous

ohhh ok so on the left side for any problem i will always decrease 1 and rise 1?

- anonymous

im trying to understand the concept so i can graph any absolute value problem

- 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

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

if it was |x| - 3, then you would move (0,0) to (0,-3)

- jim_thompson5910

but it's actually |x-3|

- anonymous

oh

- anonymous

would i be able to just plug this in to my calculator ?

- jim_thompson5910

yes you should be able to

- jim_thompson5910

do you know how to graph absolute value equations?

- anonymous

I guess I forgot :/

- jim_thompson5910

what kind of calculator do you have

- anonymous

i just found the absolute value button on my TI-83

- anonymous

and i put it into the y= and i got what you showed me earlier

- jim_thompson5910

ok great, glad you did

- anonymous

Solve the equation.
\[4x ^{2}-6x+9=0\]

- jim_thompson5910

you can either factor or use the quadratic formula

- jim_thompson5910

oh wait, one sec

- jim_thompson5910

no you can't factor, so you'll have to use the quadratic formula

- anonymous

how do you know you cant factor this problem?

- 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

oh okay, and i can use the discriminant as a test on any problem that looks like this one?

- jim_thompson5910

yes if the discriminant is a) positive or zero and b) a perfect square
then the quadratic can be factored

- anonymous

ohhhhh okay

- anonymous

solve the system of linear equations using any method.
2x+3y=47
7x-8y=-2
2x-y+3z=-19

- jim_thompson5910

solve the first two equations for x and y
once you have x and y, use them to find z

- anonymous

im not sure how i ended up with 111 for y... the answers say its wrong

- jim_thompson5910

no y should be 9, try it again

- anonymous

i did elimination with the first two equations

- anonymous

to get them to equal to 14 and -14 so they can cancel out but i keep getting 111

- 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

Oh! I understand now, thank you so much :)

- jim_thompson5910

you're welcome

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