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hichkok12 [17]
3 years ago
6

You

Mathematics
2 answers:
Eva8 [605]3 years ago
6 0

.25(4) = 1 gallon

.25(8) = 2 gallons

.25(12) = 3 gallons

.25(16) = 4 gallons

.25(20) = 5 gallons

it would take 20 minutes to remove half the water (aka 5 gallons)

Katarina [22]3 years ago
4 0

Answer:

20

Step-by-step explanation:

look up

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If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
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\bf f(x)=y=2x+sin(x)
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inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
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\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
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1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
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now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
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-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
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