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miskamm [114]
3 years ago
7

How are the functions f(x) = 16* and 19(x) = 161/2 related?

Mathematics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

F(x) = 16^x = 8^2x

g(x) = 16^1/2x = 8^x

So you can see the relationship. They are all of the type a^bx form, with same base of a, f(x) will have larger increasing rate.

Step-by-step explanation:

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Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

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\bf \textit{hyperbolas, vertical traverse axis }&#10;\\\\&#10;\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h,  k\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2 + b ^2}\\&#10;asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=2\\&#10;k=-3\\&#10;a=6\\&#10;c=2\sqrt{10}&#10;\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2&#10;\\\\\\&#10;4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2&#10;\\\\\\&#10;\sqrt{4}=b\implies 2=b\\\\&#10;-------------------------------\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

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