anybody in geomtry in ga virtual school?

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anybody in geomtry in ga virtual school?

Mathematics
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I'm not personally, but if you asked your question, I'm sure someone would be able to help you.
i need help with this
@freckles , @mathmate , or @Michele_Laino would probably be able to help you with that. I haven't gotten to that yet.

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@Michele_Laino @mathmate @freckles could any one of yall help me
oh are you not in 10th grade?
question #1a the requested inequality is: \[ \Large - 12{x^2} + 960x \geqslant 18000\]
thank you but i need help pn 3a 3b
@Michele_Laino could you help me
question #3a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 18432\]
dividing both sides by 12, we can rewrite that inequality as follows: \[\Large {x^2} - 80x + 1536 \geqslant 0\]
so thats the inequality?
would the answer be
the correspnding equation is: \[\Large 12{x^2} - 960x + 18432 = 0\] whereas the critical values are: x=32, and x=48
finally the solution to that inequality is: \[\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x\]
ok thank you so for 3a where as it says write an inequailty when she will not be taxes will it be >/ 50 because on the graph
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ok!
im asking is that correct?
the requested inequality is \[\Large x \geqslant 48\]
ok thank you. could you help me on more or are you stressed out?
I can help you! :)
ok thankyou here are the other ones
question #4a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 14400\]
how would you do b?
here is the graph:
THANK YOU ! so i would shade below the graph correct?
yes! correct!
ok for E it says are any of the values not in the orginal domain explain your answer and write your final solution as inequality.
yes all the values x, such that: \[\Large 0 \leqslant x < 30\] don't belong to the original domain of our function: \[\Large y=- 12{x^2} + 960x\]
so the solution is: \[\Large x \geqslant 60\]
wait so how would x> 60 belong in the equation ?
since the intersection between the graph: \[y = - 12{x^2} + 960x\] and the line: \[y = 14400\] occurs at x=20 and x=60. Now all x-values, such that 0/ 60
ok thank you what about these
question #5 as we can see from the drawing below, our solution is: \[x \leqslant 1 \cup x \geqslant 5\]
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what does u stand for?
it is the union of sets
do i put and or or ? or u
that writing is equivalent to this one: \[( - \infty ,1] \cup [5, + \infty )\] you can put "or"
Question #6 as we can see from the drawing below, our solution is: \[ - 1 \leqslant x \leqslant 5\]
1 Attachment
Question #7 here you have to solve the quadratic equation associated to your inequality, first. For example, if we consider case a), we get: \[{x^2} - 4x + 3 < 0\] so you have to solve this equation: \[{x^2} - 4x + 3 = 0\] what do you get?
x=3,1?
ok!
so we can write the solution as an inequality below: \[1 < x < 3\]
shaded to the left?
it is: |dw:1434396507609:dw| and x=1 , x=3 don't belong to the solution interval
case b) here after a simplification, we get: \[3{x^2} - 5x - 12 > 0\] so we have to solve this equation: \[3{x^2} - 5x - 12 = 0\] what do you get?
1.93?
are you sure?
please wait a moment
ya...
I got: x=3, and x=-4/3 so our solution is: \[x < - 4/3,\;x > 3\] |dw:1434399244525:dw|
x=3, and x=-4/3 don't belong to the solution interval
case c) we can write: \[{x^2} + 6x + 9 \geqslant 0\] now we have to nothe that: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\]
so our inequality is checked for all real values of x: |dw:1434399797411:dw| in x=-3 we have: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0\]
i got x>/ -3?
ok thank you for 1d on paula peaches i got 30
the inequality: \[{x^2} + 6x + 9 \geqslant 0\] is always true, and the equal sign holds, when x=-3
30/60
more precisely: 60
for 2b i got x would equal 50, 30 and c im a little lost on thats the last two
there we have to solve this inequality: \[ - 12{x^2} + 960x \geqslant 18000\]
which, after a simplification, becomes: \[{x^2} - 80x + 1500 \leqslant 0\]
now the roots of the associated quadratic equation: \[{x^2} - 80x + 1500 = 0\] are: x=50 and x=30
so the requested solution is: \[30 \leqslant x \leqslant 50\]
so we have the subsequent graph: |dw:1434401181442:dw|
thank you so much for the help you answered my questions fully

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