Question

7 part question

Answer 1

so the key for these is to graph the equations as if the >, <, etc. signs were equal signs first

Answer 2

i already know part IV

Answer 3

then you would use the inequality symbols (>, <. etc.) to determine the line style and shaded area

Answer 4

okay so i would graph y=1/2x+6

Answer 5

yes

Answer 6

https://www.desmos.com/calculator

Answer 7

good, so y > (1/2)x + 6

if you have > or < you use a dashed line to indicate that the solutions do not include the line

Answer 8

[this is just an example of how it looks, this is not the solution]

Answer 9

the > greater than symbol means you would shade all the area above the line

Answer 10

wait but what line is this

Answer 11

which line am i shading above

Answer 12

are we still on part I

Answer 13

oh whoops it's y < (1/2)x + 6 so it would be shading under the line y = (1/2)x + 6

Answer 14

anyway, I think it might be less overwhelming if you only consider one line at a time

Answer 15

but we are on part I still right

Answer 16

yes

Answer 17

okay so i would shade it like this right?

Answer 18

yes

Answer 19

but make sure you understand why it's being shaded the way it is

Answer 20

because of the < , > ssigns right?

Answer 21

I would suggest just graphing one inequality at a time to see what's going on

Answer 22

yes, that's correct

Answer 23

okay but woud i shae exactly like the graph

Answer 24

yes

Answer 25

no i would shade under the y 1/2 line because that is where all the points are common

Answer 26

right, you would shade the area where all of the colors overlap

Answer 27

okay now for the part on whether the lines i graphed they would be solid?

Answer 28

for any inequalities with > or < use a dashed line

Answer 29

for any inequalities with \[\le or \ge\] you would use a solid line

Answer 30

so all of them solid except for the y1/2 line

Answer 31

yes

Answer 32

did i shade it wrong? :S

Answer 33

for part I it is only asking for x = 0 and x = 12

Answer 34

just the lines

Answer 35

ugh im confused

Answer 36

yes, it's just asking you to graph x = 0 and x = 12 as if they were normal lines

Answer 37

okay so this is part I

Answer 38

right that's it

Answer 39

okay now for part II

Answer 40

would it be y1/2 dashed line and the rest solid

Answer 41

you have only graphed x = 0 and x = 12 up to this point

so you would only answer the question for those two lines

Answer 42

solid?

Answer 43

good, so you would answer: the two lines in part I are solid

Answer 44

now for part III

Answer 45

i would graph what is given and just see where the colors overlap right?

Answer 46

yes. keep in mind you are only graphing x >= 0 and x <= 12 as stated in the directions.

Answer 47

good (make sure you only shade the part where it overlaps)

Answer 48

good

Answer 49

parts IV and V are pretty self explanatory just follow the directions

Answer 50

V the line should be dashed

Answer 51

and the shading is below

Answer 52

good

Answer 53

now for part VI

Answer 54

x + 3y = 12

1. find the y-intercept by finding y when x = 0
2. find the x-intercept by finding x when y = 0
3. connect the points from 1 and 2

Answer 55

y=4
x=12

Answer 56

good now connect (0,4) and (12,0) in a line

Answer 57

okay

Answer 58

part VII

Answer 59

so to determine the direction of shading I would suggest re-writing the equation in terms of y

Answer 60

so put in in y __ mx + b form

Answer 61

y=3x+12

Answer 62

try one more time

Answer 63

y=12x+3

Answer 64

let's start with x + 3y = 12

solve for y

Answer 65

y=x/3-4

Answer 66

good, so adding back the inequality sign we have

y >= (x/3) - 4

which direction do we shade? above or below?

Answer 67

above

Answer 68

good, so shade above and that's part VII solved

Answer 69

okay part VIII

Answer 70

now you just put together all of the graphs you drew

Answer 71

the final result should be like this, with that middle triangle-y section shaded in

Answer 72

okay thank you, i am in the middle of something i will need to come back to this something popped up

Answer 73

I am back

Answer 74

@Vocaloid

Answer 75

Okay so I just draw the lines of the equations given for VIII and shade in the region you showed

Answer 76

May I see what you've drawn?

Answer 77

nothing yet.. I am going to draw it now, I had to leave urgently

Answer 78

so didn't get to draw anything for VIII

Answer 79

oh, ok

Answer 80

this is my drawing

Answer 81

check the slope on that dotted line (the slope should be 1/2)

Answer 82

also the y-intercept on the solid line is incorrect (should be 0,4)

Answer 83

the line that goes across the dotted line

Answer 84

okay fixed that

Answer 85

sorry I was afk can I see what you have now?

Answer 86

it is alright

Answer 87

awesome, now just shade in the big triangle (don't include the left tip part since that's to the left of x = 0

Answer 88

just that triangle in the smack down middle right

Answer 89

yes