Mathematics
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OpenStudy (anonymous):
For the graphed function f(x) = (4)x - 1 + 2, calculate the average rate of change from x = 2 to x = 4.
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OpenStudy (anonymous):
OpenStudy (anonymous):
@johnweldon1993
OpenStudy (anonymous):
@ganeshie8 ...
ganeshie8 (ganeshie8):
the asked question and the picture u attached are not matching
ganeshie8 (ganeshie8):
check once..
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OpenStudy (anonymous):
okay sorry, just click the picture then...
OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
how?
ganeshie8 (ganeshie8):
average rate of change from x = 2 to x = 4 is \(\large \dfrac{f(4) - f(2)}{4-2}\)
ganeshie8 (ganeshie8):
first find f(4) and f(2)
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ganeshie8 (ganeshie8):
\(\large f(x) = 4^{x-1}+2\)
ganeshie8 (ganeshie8):
plugin x = 4 for f(4)
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
would 4 be the exponent?
ganeshie8 (ganeshie8):
\(\large f(4) = 4^{4-1}+2 = 4^3 + 2 = 64 + 2 = 66\)
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ganeshie8 (ganeshie8):
similarly find f(2) and plug them in the average rate of change formula
OpenStudy (anonymous):
okay, would i subtract the answers?
ganeshie8 (ganeshie8):
yes, and then divide by 4-2
ganeshie8 (ganeshie8):
\(\large f(2) = 4^{2-1}+2 = 4^2 + 2 = 4 + 2 = 6\)
OpenStudy (anonymous):
30
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OpenStudy (anonymous):
right?
ganeshie8 (ganeshie8):
30 is right !
OpenStudy (anonymous):
Thank you :) can you help me with another
OpenStudy (anonymous):
@ganeshie8
ganeshie8 (ganeshie8):
sure, ask
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OpenStudy (anonymous):
ganeshie8 (ganeshie8):
\[\large \log_{2x-5} 25 = 2\]
\[\large \log_{2x-5} 5^2 = 2\]
\[\large 2\log_{2x-5} 5 = 2\]
\[\large \log_{2x-5} 5 = 1\]
ganeshie8 (ganeshie8):
change it to exponent form
ganeshie8 (ganeshie8):
\[\large 5 = (2x-5)^1\]
ganeshie8 (ganeshie8):
solve \(x\)
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OpenStudy (anonymous):
I got 5... is that right
ganeshie8 (ganeshie8):
5 is right !!
OpenStudy (anonymous):
wow thanks! i have 2 more is that okay?
OpenStudy (anonymous):
ganeshie8 (ganeshie8):
wat do u think the equation is
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OpenStudy (anonymous):
the first one?
ganeshie8 (ganeshie8):
right !
OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
that's the right answer?!
OpenStudy (anonymous):
i have one more!
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OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
i think it's the second one actually.
OpenStudy (anonymous):
@ankit042 the most recent one!