anybody in geomtry in ga virtual school?

- ashontae19

anybody in geomtry in ga virtual school?

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- chestercat

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- k_lynn

I'm not personally, but if you asked your question, I'm sure someone would be able to help you.

- ashontae19

i need help with this

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- 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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## More answers

- ashontae19

@Michele_Laino @mathmate @freckles could any one of yall help me

- ashontae19

oh are you not in 10th grade?

- Michele_Laino

question #1a
the requested inequality is:
\[ \Large - 12{x^2} + 960x \geqslant 18000\]

- ashontae19

thank you but i need help pn 3a 3b

- ashontae19

@Michele_Laino could you help me

- Michele_Laino

question #3a
the requested inequality is:
\[\Large - 12{x^2} + 960x \leqslant 18432\]

- Michele_Laino

dividing both sides by 12, we can rewrite that inequality as follows:
\[\Large {x^2} - 80x + 1536 \geqslant 0\]

- ashontae19

so thats the inequality?

- ashontae19

would the answer be

- 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

finally the solution to that inequality is:
\[\Large 0 \leqslant x \leqslant 32 \cup 48 \leqslant x\]

- 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

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- Michele_Laino

ok!

- ashontae19

im asking is that correct?

- Michele_Laino

the requested inequality is
\[\Large x \geqslant 48\]

- ashontae19

ok thank you. could you help me on more or are you stressed out?

- Michele_Laino

I can help you! :)

- ashontae19

ok thankyou here are the other ones

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- Michele_Laino

question #4a
the requested inequality is:
\[\Large - 12{x^2} + 960x \leqslant 14400\]

- ashontae19

how would you do b?

- Michele_Laino

here is the graph:

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- ashontae19

THANK YOU ! so i would shade below the graph correct?

- Michele_Laino

yes! correct!

- Michele_Laino

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- 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

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

so the solution is:
\[\Large x \geqslant 60\]

- ashontae19

wait so how would x> 60 belong in the equation ?

- 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

ok thank you what about these

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- Michele_Laino

question #5
as we can see from the drawing below, our solution is:
\[x \leqslant 1 \cup x \geqslant 5\]

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- ashontae19

what does u stand for?

- Michele_Laino

it is the union of sets

- ashontae19

do i put and or or ? or u

- Michele_Laino

that writing is equivalent to this one:
\[( - \infty ,1] \cup [5, + \infty )\]
you can put "or"

- Michele_Laino

Question #6
as we can see from the drawing below, our solution is:
\[ - 1 \leqslant x \leqslant 5\]

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- 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

x=3,1?

- Michele_Laino

ok!

- Michele_Laino

so we can write the solution as an inequality below:
\[1 < x < 3\]

- ashontae19

shaded to the left?

- Michele_Laino

it is:
|dw:1434396507609:dw|
and x=1 , x=3 don't belong to the solution interval

- 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

1.93?

- Michele_Laino

are you sure?

- Michele_Laino

please wait a moment

- ashontae19

ya...

- Michele_Laino

I got:
x=3, and x=-4/3
so our solution is:
\[x < - 4/3,\;x > 3\]
|dw:1434399244525:dw|

- Michele_Laino

x=3, and x=-4/3 don't belong to the solution interval

- 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

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

i got x>/ -3?

- ashontae19

ok thank you for 1d on paula peaches i got 30

- Michele_Laino

the inequality:
\[{x^2} + 6x + 9 \geqslant 0\] is always true, and the equal sign holds, when x=-3

- Michele_Laino

30/60

- Michele_Laino

more precisely:
60

- ashontae19

for 2b i got x would equal 50, 30 and c im a little lost on thats the last two

- Michele_Laino

there we have to solve this inequality:
\[ - 12{x^2} + 960x \geqslant 18000\]

- Michele_Laino

which, after a simplification, becomes:
\[{x^2} - 80x + 1500 \leqslant 0\]

- Michele_Laino

now the roots of the associated quadratic equation:
\[{x^2} - 80x + 1500 = 0\]
are:
x=50 and x=30

- Michele_Laino

so the requested solution is:
\[30 \leqslant x \leqslant 50\]

- Michele_Laino

so we have the subsequent graph:
|dw:1434401181442:dw|

- ashontae19

thank you so much for the help you answered my questions fully

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