ashontae19
  • ashontae19
anybody in geomtry in ga virtual school?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
k_lynn
  • k_lynn
I'm not personally, but if you asked your question, I'm sure someone would be able to help you.
ashontae19
  • ashontae19
i need help with this
k_lynn
  • k_lynn
@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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ashontae19
  • ashontae19
@Michele_Laino @mathmate @freckles could any one of yall help me
ashontae19
  • ashontae19
oh are you not in 10th grade?
Michele_Laino
  • Michele_Laino
question #1a the requested inequality is: \[ \Large - 12{x^2} + 960x \geqslant 18000\]
ashontae19
  • ashontae19
thank you but i need help pn 3a 3b
ashontae19
  • ashontae19
@Michele_Laino could you help me
Michele_Laino
  • Michele_Laino
question #3a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 18432\]
Michele_Laino
  • Michele_Laino
dividing both sides by 12, we can rewrite that inequality as follows: \[\Large {x^2} - 80x + 1536 \geqslant 0\]
ashontae19
  • ashontae19
so thats the inequality?
ashontae19
  • ashontae19
would the answer be
Michele_Laino
  • Michele_Laino
the correspnding equation is: \[\Large 12{x^2} - 960x + 18432 = 0\] whereas the critical values are: x=32, and x=48
Michele_Laino
  • Michele_Laino
finally the solution to that inequality is: \[\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x\]
ashontae19
  • ashontae19
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
1 Attachment
Michele_Laino
  • Michele_Laino
ok!
ashontae19
  • ashontae19
im asking is that correct?
Michele_Laino
  • Michele_Laino
the requested inequality is \[\Large x \geqslant 48\]
ashontae19
  • ashontae19
ok thank you. could you help me on more or are you stressed out?
Michele_Laino
  • Michele_Laino
I can help you! :)
ashontae19
  • ashontae19
ok thankyou here are the other ones
Michele_Laino
  • Michele_Laino
question #4a the requested inequality is: \[\Large - 12{x^2} + 960x \leqslant 14400\]
ashontae19
  • ashontae19
how would you do b?
Michele_Laino
  • Michele_Laino
here is the graph:
ashontae19
  • ashontae19
THANK YOU ! so i would shade below the graph correct?
Michele_Laino
  • Michele_Laino
yes! correct!
Michele_Laino
  • Michele_Laino
ashontae19
  • ashontae19
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.
Michele_Laino
  • Michele_Laino
yes all the values x, such that: \[\Large 0 \leqslant x < 30\] don't belong to the original domain of our function: \[\Large y=- 12{x^2} + 960x\]
Michele_Laino
  • Michele_Laino
so the solution is: \[\Large x \geqslant 60\]
ashontae19
  • ashontae19
wait so how would x> 60 belong in the equation ?
Michele_Laino
  • Michele_Laino
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
ashontae19
  • ashontae19
ok thank you what about these
Michele_Laino
  • Michele_Laino
question #5 as we can see from the drawing below, our solution is: \[x \leqslant 1 \cup x \geqslant 5\]
1 Attachment
ashontae19
  • ashontae19
what does u stand for?
Michele_Laino
  • Michele_Laino
it is the union of sets
ashontae19
  • ashontae19
do i put and or or ? or u
Michele_Laino
  • Michele_Laino
that writing is equivalent to this one: \[( - \infty ,1] \cup [5, + \infty )\] you can put "or"
Michele_Laino
  • Michele_Laino
Question #6 as we can see from the drawing below, our solution is: \[ - 1 \leqslant x \leqslant 5\]
1 Attachment
Michele_Laino
  • Michele_Laino
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?
ashontae19
  • ashontae19
x=3,1?
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
so we can write the solution as an inequality below: \[1 < x < 3\]
ashontae19
  • ashontae19
shaded to the left?
Michele_Laino
  • Michele_Laino
it is: |dw:1434396507609:dw| and x=1 , x=3 don't belong to the solution interval
Michele_Laino
  • Michele_Laino
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?
ashontae19
  • ashontae19
1.93?
Michele_Laino
  • Michele_Laino
are you sure?
Michele_Laino
  • Michele_Laino
please wait a moment
ashontae19
  • ashontae19
ya...
Michele_Laino
  • Michele_Laino
I got: x=3, and x=-4/3 so our solution is: \[x < - 4/3,\;x > 3\] |dw:1434399244525:dw|
Michele_Laino
  • Michele_Laino
x=3, and x=-4/3 don't belong to the solution interval
Michele_Laino
  • Michele_Laino
case c) we can write: \[{x^2} + 6x + 9 \geqslant 0\] now we have to nothe that: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\]
Michele_Laino
  • Michele_Laino
so our inequality is checked for all real values of x: |dw:1434399797411:dw| in x=-3 we have: \[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0\]
ashontae19
  • ashontae19
i got x>/ -3?
ashontae19
  • ashontae19
ok thank you for 1d on paula peaches i got 30
Michele_Laino
  • Michele_Laino
the inequality: \[{x^2} + 6x + 9 \geqslant 0\] is always true, and the equal sign holds, when x=-3
Michele_Laino
  • Michele_Laino
30/60
Michele_Laino
  • Michele_Laino
more precisely: 60
ashontae19
  • ashontae19
for 2b i got x would equal 50, 30 and c im a little lost on thats the last two
Michele_Laino
  • Michele_Laino
there we have to solve this inequality: \[ - 12{x^2} + 960x \geqslant 18000\]
Michele_Laino
  • Michele_Laino
which, after a simplification, becomes: \[{x^2} - 80x + 1500 \leqslant 0\]
Michele_Laino
  • Michele_Laino
now the roots of the associated quadratic equation: \[{x^2} - 80x + 1500 = 0\] are: x=50 and x=30
Michele_Laino
  • Michele_Laino
so the requested solution is: \[30 \leqslant x \leqslant 50\]
Michele_Laino
  • Michele_Laino
so we have the subsequent graph: |dw:1434401181442:dw|
ashontae19
  • ashontae19
thank you so much for the help you answered my questions fully

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