3×4×100±67≤65║556∪120D⊥p=13⇆87∞44f1······ HELP PLEASEE WHAT IS THIS????

Mathematics

1 answer:

kkurt [141]2 years ago

8 0

Answer: That is most likely an unsolvable question.

Step-by-step explanation: PEMDAS

3×4×100±67≤65 Doesn’t work.

3x4= 12 ~ 12x100=1200 ~ 1200+67=1267 ~ 1267<65

556∪120D⊥p=13⇆87∞44f1······ Also clearly isn’t a math question

I am not sure if this helps. But I really did look into it and found nothing. I am So sorry to disappoint.

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olchik [2.2K]

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PLEASE HELP!!!! Which of these pair of functions are inverse functions?

Mamont248 [21]

Answer:

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;

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