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vichka [17]
3 years ago
9

Please help! I will give brainlist, if I can

Mathematics
2 answers:
Dmitry [639]3 years ago
6 0

your awnser is C

Step-by-step explanation:

cause you'll need to multiply 1.05 × 15 then divide by 100

Shtirlitz [24]3 years ago
3 0
It is B glad I could help!
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A large bin of mixed nuts contains peanuts, cashews, almonds, and walnuts. Sheila scoops out a bowl full of nuts and counts how
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Assuming that the ratios stay the same, the answer is going to be 6 because the total number of nuts drawn originally was 48, and 9 were cashews. To maintain this ratio when 32 nuts were drawn, the number of cashews will be 6.
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Find the distance between each pair if points
sukhopar [10]

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Which statements are true based on the diagram? Select three options.
Aneli [31]

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Are Correct.

Step-by-step explanation:

5 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\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}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

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
\\\\\\
-----------------------------\\\\
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
5 0
3 years ago
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