Question

anybody in geomtry in ga virtual school?

Answer 1

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

Answer 2

i need help with this

Answer 3

@freckles , @mathmate , or @Michele_Laino would probably be able to help you with that. I haven't gotten to that yet.

Answer 4

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

Answer 5

oh are you not in 10th grade?

Answer 6

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

Answer 7

thank you but i need help pn 3a 3b

Answer 8

@Michele_Laino could you help me

Answer 9

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

Answer 10

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

Answer 11

so thats the inequality?

Answer 12

would the answer be </ 32 and </ 48?

Answer 13

the correspnding equation is:
\[\Large 12{x^2} - 960x + 18432 = 0\]
whereas the critical values are:
x=32, and x=48

Answer 14

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

Answer 15

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

Answer 16

ok!

Answer 17

im asking is that correct?

Answer 18

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

Answer 19

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

Answer 20

I can help you! :)

Answer 21

ok thankyou here are the other ones

Answer 22

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

Answer 23

how would you do b?

Answer 24

here is the graph:

Answer 25

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

Answer 26

yes! correct!

Answer 27

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.

Answer 28

yes all the values x, such that:
\[\Large 0 \leqslant x < 30\]
don't belong to the original domain of our function:
\[\Large y=- 12{x^2} + 960x\]

Answer 29

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

Answer 30

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

Answer 31

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</ x</20 don't belong to the domain of our function, so we have towrite x>/ 60

Answer 32

ok thank you what about these

Answer 33

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

Answer 34

what does u stand for?

Answer 35

it is the union of sets

Answer 36

do i put and or or ? or u

Answer 37

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

Answer 38

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

Answer 39

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?

Answer 40

x=3,1?

Answer 41

ok!

Answer 42

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

Answer 43

shaded to the left?

Answer 44

it is:

and x=1 , x=3 don't belong to the solution interval

Answer 45

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?

Answer 46

1.93?

Answer 47

are you sure?

Answer 48

please wait a moment

Answer 49

ya...

Answer 50

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

Answer 51

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

Answer 52

case c)
we can write:
\[{x^2} + 6x + 9 \geqslant 0\]
now we have to nothe that:
\[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\]

Answer 53

so our inequality is checked for all real values of x:

in x=-3 we have:
\[{x^2} + 6x + 9 = {\left( {x + 3} \right)^2} = 0\]