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3 years ago

Please help!

Mathematics

1 answer:

zhuklara [117]3 years ago

6 0

Exponential Growth Formula: $y = b(1+r)^t$

$b = 315\\r = 0.06\\t = 7\\\\y = 315(1+0.06)^7\\y = 315(1.06)^7\\y = 315(1.50363)\\y = 473.64345\\y = 473.64$

After 7 years, she will make $473.64

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Option B and C are correct.

Step-by-step explanation:

Inverse function: If both the domain and the range are R for a function f(x), and if f(x) has an inverse g(x) then:

$f(g(x)) = g(f(x)) = x$ for every x∈R.

Let $f(x) = \frac{1}{2}(\ln(\frac{x}{2}) -1)$ and $g(x) = 2e^{2x+1}$

Use logarithmic rules:

$ln e^a = a$

$e^{lnx} = x$

$\ln a^b = b\ln a$

then, by definition;

$f(g(x)) = f(2e^{2x+1}) =\frac{1}{2}(\ln(\frac{2e^{2x+1}}{2})-1)$ = $\frac{1}{2}(\ln(e^{2x+1}}){-1) = \frac{1}{2} (2x+1-1) =\frac{1}{2}(2x) = x$

$g(f(x)) = g(\frac{1}{2}(\ln(\frac{x}{2}) -1)) = 2e^{2({\frac{1}{2}(\ln(\frac{x}{2}) -1})+1$ $2e^{(\ln(\frac{x}{2}) -1+1}=2e^{\ln(\frac{x}{2})} =2\cdot \frac{x}{2} = x$

Similarly;

for $f(x) = \frac{4 \ln(x^2)}{e^2}$ and $g(x) = e^{\frac{e^2 \cdot x}{8} }$

then, by definition;

$f(g(x)) = f(e^{\frac{e^2 \cdot x}{8}}) =\frac{4 \ln {(\frac{e^2 \cdot x}{8})^2}}{e^2}$ = $\frac{8 \ln {(\frac{e^2 \cdot x}{8})}}{e^2} =\frac{8\frac{e^2\cdot x}{8} }{e^2}=\frac{8e^2 \cdot x}{8e^2}=x$