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ashontae19
 one year ago
anybody in geomtry in ga virtual school?
ashontae19
 one year ago
anybody in geomtry in ga virtual school?

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k_lynn
 one year ago
Best ResponseYou've already chosen the best response.0I'm not personally, but if you asked your question, I'm sure someone would be able to help you.

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0i need help with this

k_lynn
 one year ago
Best ResponseYou've already chosen the best response.0@freckles , @mathmate , or @Michele_Laino would probably be able to help you with that. I haven't gotten to that yet.

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino @mathmate @freckles could any one of yall help me

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0oh are you not in 10th grade?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2question #1a the requested inequality is: \[ \Large  12{x^2} + 960x \geqslant 18000\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0thank you but i need help pn 3a 3b

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino could you help me

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2question #3a the requested inequality is: \[\Large  12{x^2} + 960x \leqslant 18432\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dividing both sides by 12, we can rewrite that inequality as follows: \[\Large {x^2}  80x + 1536 \geqslant 0\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0so thats the inequality?

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0would the answer be </ 32 and </ 48?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the correspnding equation is: \[\Large 12{x^2}  960x + 18432 = 0\] whereas the critical values are: x=32, and x=48

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2finally the solution to that inequality is: \[\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0im asking is that correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the requested inequality is \[\Large x \geqslant 48\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you. could you help me on more or are you stressed out?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I can help you! :)

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok thankyou here are the other ones

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2question #4a the requested inequality is: \[\Large  12{x^2} + 960x \leqslant 14400\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0how would you do b?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the graph:

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0THANK YOU ! so i would shade below the graph correct?

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok for E it says are any of the values not in the orginal domain explain your answer and write your final solution as inequality.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes all the values x, such that: \[\Large 0 \leqslant x < 30\] don't belong to the original domain of our function: \[\Large y= 12{x^2} + 960x\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the solution is: \[\Large x \geqslant 60\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0wait so how would x> 60 belong in the equation ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since the intersection between the graph: \[y =  12{x^2} + 960x\] and the line: \[y = 14400\] occurs at x=20 and x=60. Now all xvalues, such that 0</ x</20 don't belong to the domain of our function, so we have towrite x>/ 60

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you what about these

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2question #5 as we can see from the drawing below, our solution is: \[x \leqslant 1 \cup x \geqslant 5\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0what does u stand for?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is the union of sets

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0do i put and or or ? or u

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2that writing is equivalent to this one: \[(  \infty ,1] \cup [5, + \infty )\] you can put "or"

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Question #6 as we can see from the drawing below, our solution is: \[  1 \leqslant x \leqslant 5\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Question #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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write the solution as an inequality below: \[1 < x < 3\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0shaded to the left?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is: dw:1434396507609:dw and x=1 , x=3 don't belong to the solution interval

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2case 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please wait a moment

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I got: x=3, and x=4/3 so our solution is: \[x <  4/3,\;x > 3\] dw:1434399244525:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2x=3, and x=4/3 don't belong to the solution interval

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2case c) we can write: \[{x^2} + 6x + 9 \geqslant 0\] now we have to nothe that: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so our inequality is checked for all real values of x: dw:1434399797411:dw in x=3 we have: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0\]

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you for 1d on paula peaches i got 30 </ x</ 50 is that correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the inequality: \[{x^2} + 6x + 9 \geqslant 0\] is always true, and the equal sign holds, when x=3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.230</x</50 is not correct, since the production of peaches can not be higher than 14400, so the solution is x>/60

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2more precisely: 60</x</80

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0for 2b i got x would equal 50, 30 and c im a little lost on thats the last two

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2there we have to solve this inequality: \[  12{x^2} + 960x \geqslant 18000\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2which, after a simplification, becomes: \[{x^2}  80x + 1500 \leqslant 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now the roots of the associated quadratic equation: \[{x^2}  80x + 1500 = 0\] are: x=50 and x=30

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the requested solution is: \[30 \leqslant x \leqslant 50\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we have the subsequent graph: dw:1434401181442:dw

ashontae19
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much for the help you answered my questions fully
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