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Firdavs [7]
4 years ago
12

Solve for a. 2(a+4)+6a=48

Mathematics
1 answer:
butalik [34]4 years ago
5 0
2(a+4)+6a=48 2a+8+6a=48 8a+8=48 -8 -8 8a=40 a= 5 Copy it exactly how you see it
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Find the sum of the series: .25+.125+.0625+....
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Hope it will help you!

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Which of these true conditional statements has a false converse?
cestrela7 [59]
Correct answer is A. For example 50/2= 25 which is odd
4 0
4 years ago
Can u help me with 1/2 plz
vagabundo [1.1K]

Answer:

1.) Yes, it is a function

2.) No, it is not a function.

Step-by-step explanation:

1.) It is a function because none of the numbers in the x-axis repeat.

2.) It is not a function because the numbers in the x-axis repeats 0 three times.

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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