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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

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

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 \[f(x) = 0.1(2)^x~~~~\text{or}~~~f(x) = 0.1(2)x\]

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?

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?

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?

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

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