PLEASE HELP!! (fan+medal)
Two functions are shown in the table below.
Function 1 2 3 4 5 6
f(x) = −x2 + 4x + 12
g(x) = −x + 6
Complete the table on your own paper, then select the value that is a solution to f(x) = g(x).
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OpenStudy (anonymous):
x = 2
x = 3
x = 5
x = 6
OpenStudy (anonymous):
@pooja195 @TheSmartOne @texaschic101 @HelpABoiOut
OpenStudy (anonymous):
ok that was wrong \[f(x)=-x^2+4x+12\\
f(1)=1^2+4\times 1+12=15\]
OpenStudy (anonymous):
\[g(x)=-x+6\\
g(1)=-1+6=5\]
OpenStudy (anonymous):
Oh I understand what you did there
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OpenStudy (anonymous):
you gotta repeat this for 2, 3, 4, 5 and 6
OpenStudy (anonymous):
I think it is x=3 am I right?
OpenStudy (anonymous):
\[f(x)=-x^2+4x+12\\
f(2)=-2^2+4\times 2+12=16\]
OpenStudy (anonymous):
oh you mean for where they are equal?
OpenStudy (anonymous):
wait so f(x) has to be the same as g(x) and that would be the answer?
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OpenStudy (anonymous):
yes but they are not the same when \(x=3\)
OpenStudy (anonymous):
Oh
OpenStudy (anonymous):
I know it is not 5 because I just did it with 5
OpenStudy (anonymous):
So it is 2 or 6
OpenStudy (anonymous):
\[g(3)=3\\
f(3)=15\] not equal
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OpenStudy (anonymous):
Yeah I messed up
OpenStudy (anonymous):
f(2) = 24
OpenStudy (anonymous):
g(2) = 4
OpenStudy (anonymous):
you are making some sort of error computing \(f(x)\)
OpenStudy (anonymous):
\[f(2)=-2^2+4\times 2+12\]
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OpenStudy (anonymous):
i see the error
OpenStudy (anonymous):
\[-2^2=-2\times 2=-4\] not \((-2)^2=4\)
OpenStudy (anonymous):
Ooooh
OpenStudy (anonymous):
so \[f(2)=-2^2+4\times 2+12=-4+8+12=16\]
OpenStudy (anonymous):
still not equal to \(g(2)=4\) though `
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OpenStudy (anonymous):
then it has to be x=6
OpenStudy (anonymous):
f(6) = 0
OpenStudy (anonymous):
g(6) = 0
OpenStudy (anonymous):
bingo
OpenStudy (anonymous):
wool thank you soooo much
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OpenStudy (anonymous):
yw
OpenStudy (anonymous):
can you help me with another one?
OpenStudy (anonymous):
sure if it is quick
OpenStudy (anonymous):
I will leave testimonial fan and medal :)
OpenStudy (anonymous):
The first-, second-, and third-year enrollment values for a technical school are shown in the table below.
Enrollment at a Technical School
Year (x) First Year f(x) Second Year s(x) Third Year t(x)
2009 785 756 756
2010 740 785 740
2011 690 710 781
2012 732 732 710
2013 781 755 800
Which of the following statements is true based on the data in the table?
The solution to f(x) = s(x) is x = 2,009.
The solution to f(x) = s(x) is x = 785.
The solution to f(x) = t(x) is x = 2,010.
The solution to f(x) = t(x) is x = 740.
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OpenStudy (anonymous):
this one is quick
OpenStudy (anonymous):
if i read the table correctly they are the same in the year 2012 , first and second aer both 732
OpenStudy (anonymous):
oh but that is not an option is it?q
OpenStudy (anonymous):
first is equal to the third in 2010
OpenStudy (anonymous):
so go with option C
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OpenStudy (anonymous):
Oh I get it!!! thank you
OpenStudy (anonymous):
Last one?
OpenStudy (anonymous):
you got 3 minutes
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
and this one will cost $12
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OpenStudy (anonymous):
A graph of 2 functions is shown below.
graph of function f of x equals 2 multiplied by x plus 1 and graph of function g of x equals x cubed plus 2 multiplied by x squared minus x minus 2
Which of the following is an approximate solution for f(x) = g(x)?
x = −2
x = 0
x = −1
x = 1
OpenStudy (anonymous):
Oh so never mind
OpenStudy (anonymous):
But thank you
OpenStudy (anonymous):
just kidding
OpenStudy (anonymous):
lol I was like pellettt
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OpenStudy (anonymous):
\[f(x)=2x+1\\
g(x)=x^3+2x^2-x-2\] like that?
OpenStudy (anonymous):
OpenStudy (anonymous):
you are looking for the x value of the points where the line and the curve intersect
OpenStudy (anonymous):
I guess, I don't understand things with graphs
OpenStudy (anonymous):
the one on the left, i would say the x value is about \(-2.5\)
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OpenStudy (anonymous):
you see where they cross right?
OpenStudy (anonymous):
-1
OpenStudy (anonymous):
yes they also cross close to \(x=-1\)
OpenStudy (anonymous):
and again at about \(x=1.5\) or so
OpenStudy (anonymous):
but only \(-1\) is an option, so go with that one
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