Mathematics
14 Online
OpenStudy (anonymous):
For f(x) = 0.01(2)x, find the average rate of change from x = 12 to x = 15.
3
40.96
95.57
286.72
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OpenStudy (anonymous):
@Astrophysics
OpenStudy (astrophysics):
Rate of change is just the slope
OpenStudy (anonymous):
SO,
OpenStudy (astrophysics):
\[rate~of~change = \frac{ f(b)-f(a) }{ b-a }\]
OpenStudy (astrophysics):
so find f(b) and f(a) where a = 12, and b = 15
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OpenStudy (jdoe0001):
\(\bf slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies
\cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}\qquad thus
\\ \quad \\
\cfrac{{\color{blue}{ f(15)}-{\color{blue}{ f(12)}}}}{{\color{red}{ 15-{\color{red}{ 12}}}}}\impliedby \textit{rate of change}\)
OpenStudy (astrophysics):
|dw:1436397137594:dw|
OpenStudy (astrophysics):
For a better understanding
OpenStudy (anonymous):
3?
OpenStudy (astrophysics):
Is that \[f(x) = 0.1(2)^x~~~~\text{or}~~~f(x) = 0.1(2)x\]
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OpenStudy (astrophysics):
I'm assuming it's exponent and no you're wrong
OpenStudy (astrophysics):
\[f(15) = 0.01(2)^{15}\]
OpenStudy (astrophysics):
\[f(12) = 0.01(2)^{12}\]
OpenStudy (astrophysics):
\[rate~of~change = \frac{ 0.01(2)^{15}-0.01(2)^{12} }{ 15-12 }\]
OpenStudy (anonymous):
286.72?
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OpenStudy (astrophysics):
No :\, please just use a calculator
OpenStudy (astrophysics):
You need to divide by 3
OpenStudy (anonymous):
i did
OpenStudy (anonymous):
wait let me see
OpenStudy (astrophysics):
What did you get?
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OpenStudy (anonymous):
i 95.57
OpenStudy (astrophysics):
Yup!
OpenStudy (astrophysics):
Make sure you understand what we did though
OpenStudy (astrophysics):
Once again this is just the slope, the image should show you what it means |dw:1436398295490:dw|
OpenStudy (anonymous):
oh ok thank you!!! can you help me with two more?
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OpenStudy (anonymous):
@Sarcasmus
OpenStudy (anonymous):
Which of the following options is an equivalent function to f(x) = 3(2)3x?
f(x) = 3(8)x
f(x) = 24x
f(x) = 27(8)x
f(x) = 3(8x)
OpenStudy (anonymous):
@muscrat123
OpenStudy (anonymous):
@muscrat123
OpenStudy (anonymous):
@muscrat123