Which of the following ordered pairs is not a solution to the inequality y < -x + 4?
(1, 2)
(2, 1)
(4, 0)
(3, -2)
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OpenStudy (anonymous):
just sub in all your x and y into the equations until you get something that doesn't make sense.
OpenStudy (anonymous):
how do i do that
OpenStudy (anonymous):
your points are int he form of (x,y)
OpenStudy (anonymous):
(1,2) for example x=1 and y=2
OpenStudy (anonymous):
y+x<4
just plugin the values of x&y to satisfy the above condition.
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OpenStudy (anonymous):
y<-x+4 so for the first one you would end up with 2<-1+4 which equals 2<3
OpenStudy (anonymous):
is it D
OpenStudy (anonymous):
b
OpenStudy (anonymous):
no what did you get for b as you equation?
OpenStudy (anonymous):
2 < -1 +4
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OpenStudy (anonymous):
solve the right hand side
OpenStudy (anonymous):
-1 + 4 = 3
OpenStudy (anonymous):
yup so your equation is 2<3 is this right or rwrong? this equation says that 2 is less than 3
OpenStudy (anonymous):
its right
OpenStudy (anonymous):
wait is it c because its 4 < 4
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OpenStudy (anonymous):
yup and your wuestion is asking what point makes this pint wrong.
OpenStudy (anonymous):
yup :)
OpenStudy (anonymous):
you got it
OpenStudy (anonymous):
can you help with one more?
OpenStudy (anonymous):
yea no problem
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OpenStudy (anonymous):
Part 1: Solve the compound inequality 3x + 6 > 36 and 3x + 6 54 and show all of your work. You can use <= to represent the less than or equal to symbol.
Part 2: Use complete sentences to describe the graph of the solution from Part 1.
OpenStudy (anonymous):
okay lets look at your first compound inequality. 3x+6>36
attempt to solve this , what do you get?
OpenStudy (anonymous):
well do i - 6 from both sides? or the 3?
OpenStudy (anonymous):
subtract the 6
OpenStudy (anonymous):
if you subtracted 3 you would get 3x+3>33 which is useless if your trying to get x all by itself on a side
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OpenStudy (anonymous):
from 36 and 3 right
OpenStudy (anonymous):
or from 6 because you wanna cancel it out?
OpenStudy (anonymous):
from the left side and the right side. so 3x + 6 -6>36-6
OpenStudy (anonymous):
3x>42
OpenStudy (anonymous):
you sure?
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OpenStudy (anonymous):
3x>30?
OpenStudy (anonymous):
yup. now what would be your next step if you wanted to get one side to have just x
?
OpenStudy (anonymous):
divide by 3 on both sides
OpenStudy (anonymous):
x=10
OpenStudy (anonymous):
yup and now whats your next equation?
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OpenStudy (anonymous):
oh wait x doest EQUAL to 10 but is rather ...
OpenStudy (anonymous):
so now i do the other side the same
OpenStudy (anonymous):
your equation was 3x>30 so x>10 not equal to 10
OpenStudy (anonymous):
x=16
OpenStudy (anonymous):
> yes
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OpenStudy (anonymous):
sorry
OpenStudy (anonymous):
that's okay and I cant see your second equation what was it again?
OpenStudy (anonymous):
x<16 (3x + 6 > 36 and 3x + 6 54)
OpenStudy (anonymous):
okay for your second equation all I can see is 3x + 6 54
which makes no sense to me whats your sign > < or =
OpenStudy (anonymous):
less than or equal to sorry didnt see that
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OpenStudy (anonymous):
now that i have that do i just put x<10 and x is less than or equal to 16 and thats the answer ?
OpenStudy (anonymous):
x>10
OpenStudy (anonymous):
now you have to graph it
OpenStudy (anonymous):
and in you graph your x values hve to be less than ten or greater than o equal to 16
OpenStudy (anonymous):
6 WOULD BE CLOSED CIRCLE TO THE RIGHT?
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OpenStudy (anonymous):
...where di you get 6 from?
OpenStudy (anonymous):
16
OpenStudy (anonymous):
yup then
OpenStudy (anonymous):
and 10 would closed circle to the left
OpenStudy (anonymous):
thats how i graph it right?
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OpenStudy (anonymous):
I think since your value cant be ten you would have to have an open acircle