well because f(x) = g(x)
so g(x) is x of f(x)
so you put the eqaution of g(x) into the the equation of f(x)
OpenStudy (kagıtucak):
In both of them, you just need to put g(x) instead of x in f(x).
For the first one:
f(x)=9x-2 then
f(g(x))=9(-x+3)-2
=-9x+27-2
=-9x+25
I leave the second one to you
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OpenStudy (anonymous):
yeah what kagitucak said or wrote
OpenStudy (anonymous):
i did that but got -9x-1?? and -40x-1??
OpenStudy (anonymous):
and the second one is the same step
OpenStudy (anonymous):
what did i do wrong??
OpenStudy (anonymous):
yeah because you subtracted first instead of multiplying what was in the paranthesis
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OpenStudy (anonymous):
so would it be -40x+33??
OpenStudy (anonymous):
31
OpenStudy (anonymous):
for the second one??
OpenStudy (anonymous):
i thought the first one was -9x+25
OpenStudy (anonymous):
yes you are correct
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OpenStudy (anonymous):
ok and wb the second one??
OpenStudy (anonymous):
second one would be B
OpenStudy (anonymous):
are you fuys good with sqrt and extraneous and non extraneous??
OpenStudy (anonymous):
Square root of x plus 6 − 4 = x
Square root of x plus 6 − 4 + 4 = x + 4
Square root of x plus 6 = x + 4
(Square root of x plus 6)2 = (x + 4)2
x + 6 = x2 + 8x + 16
x + 6 − 6 = x2 + 8x + 16 − 6
x = x2 + 8x + 10
x − x = x2 + 8x + 10 − x
0 = x2 + 7x + 10
0 = (x + 2)(x + 5)
x + 2 = 0 x + 5 = 0
x + 2 − 2 = 0 − 2 x + 5 − 5 = 0 − 5
x = −2 x = −5
Solutions = −2, −5
What error did Israel make?
OpenStudy (anonymous):
He subtracted 6 before subtracting x.
He added 4 before squaring both sides.
He factored x2 + 7x + 10 incorrectly.
He did not check for extraneous solutions.
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OpenStudy (anonymous):
is it d??
OpenStudy (anonymous):
hmmm
OpenStudy (anonymous):
i think its d i dont see anything else worng
OpenStudy (anonymous):
well i think that it is D too but im not sure because i hate word problems
OpenStudy (anonymous):
lol
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OpenStudy (anonymous):
wb this one??
Violet was carrying around the steps to solve Square root of 3x plus 1 = 4. On her way through the Math Factory, Violet dropped them and they mixed with other steps. Put the correct steps in order to solve Square root of 3x plus 1 = 4.
A. Subtract 1 from both sides B. Subtract 4 from both sides C. Square root both sides
D. Square both sides E. Divide by 3 from both sides F. Divide by 4 from both sides
First A, then E, then D
First D, then B, then A
First D, then A, then E
First C, then E, then A
OpenStudy (anonymous):
so the equation is\[\sqrt{3x + 1}=4\]
OpenStudy (anonymous):
if so first is D
OpenStudy (anonymous):
then A
OpenStudy (anonymous):
then E
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OpenStudy (anonymous):
sorry my comp spazed
OpenStudy (anonymous):
@ShailKumar
OpenStudy (anonymous):
i have a diff q those where answered
OpenStudy (anonymous):
Post it
OpenStudy (anonymous):
Determine the equation of the line, in standard form, that will get your spacecraft from the Launch Area to Point A. Launch Area: (1,2), Point A: (0, 3)
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OpenStudy (anonymous):
i got x+y=2
OpenStudy (anonymous):
How did you get that ?
OpenStudy (anonymous):
i got -1/1 and plugged it in to y-2=-1(x-0)
OpenStudy (anonymous):
then cpmverted
OpenStudy (anonymous):
You have plugged in wrong point. You should plug in e.g (1,2) then you should get
y-2 = -1(x-1)
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OpenStudy (anonymous):
ohhhhhhhh
OpenStudy (anonymous):
so x+y=1
OpenStudy (anonymous):
??
OpenStudy (anonymous):
Yes. Now it's correct. :)
OpenStudy (anonymous):
Determine the equation of the line, in point-slope form, that will get your spacecraft from Point A to Point B. Point A: (0,3), Point B: (-3, 0)
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OpenStudy (anonymous):
Show me your work
OpenStudy (anonymous):
y-3=-3(x-0)
OpenStudy (anonymous):
@AnswerMyQuestions
OpenStudy (anonymous):
the eqaution is simple just plug in the points
y - y2 = m(x-x2)
OpenStudy (anonymous):
Slope is 1
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OpenStudy (anonymous):
oh yeah but wouldnt it be negative??
OpenStudy (anonymous):
No
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Determine the equation of the line, in slope-intercept form, that will get your spacecraft from Point B to Point C. Point B: (-3, 0), Point C: (-1, -4)
OpenStudy (anonymous):
its the same thing
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OpenStudy (anonymous):
find the slope plug it in
OpenStudy (anonymous):
slope is y2-y1/x2-x1
OpenStudy (anonymous):
Slope intercept form is y = mx+c
OpenStudy (anonymous):
the slope is -2 right
OpenStudy (anonymous):
c is the intercept.
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OpenStudy (anonymous):
^
OpenStudy (anonymous):
Correct
OpenStudy (anonymous):
so y=-2x+??
OpenStudy (anonymous):
y-0=-2(x+3)
OpenStudy (anonymous):
y=-2x=5??
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OpenStudy (anonymous):
6
OpenStudy (anonymous):
-6
OpenStudy (anonymous):
y=-2x-6 lol
OpenStudy (anonymous):
that correct
OpenStudy (anonymous):
Yes
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OpenStudy (anonymous):
yay lol
OpenStudy (anonymous):
5) Convert the equations you arrived at in question 2 into slope-intercept form. Make sure to include all of your work.
OpenStudy (anonymous):
x+y=1
OpenStudy (anonymous):
??
OpenStudy (anonymous):
i have to put x+y=1 in to slope intercept
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OpenStudy (anonymous):
is it y=-x+1
OpenStudy (anonymous):
Correct
OpenStudy (anonymous):
last one lol
OpenStudy (anonymous):
Reflect back on this scenario and each equation you created. Would any restrictions apply to the domain and range of those equations? Explain your reasoning using complete sentences.
OpenStudy (anonymous):
wait first close this tab and create a new one please with all the scenarios together
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