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const2013 [10]
3 years ago
10

Is the equation

middle" class="latex-formula"> linear or nonlinear
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

non linear equation

Step-by-step explanation:

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Let D be a region bounded by a simple closed path C in the xy-plane. The coordinates of the centroid ( , ) of D are given below,
san4es73 [151]

Answer:

the answer is- 0.09

Step-by-step explanation:

You sound smart!

3 0
3 years ago
Explain how you can solve the following scenario in two different ways. I owe my friend $16 and my mom owes me $25. how much mon
matrenka [14]

Answer:

$9

Step-by-step explanation:

Given  I owe my friend $16 and my mom owes me $25

step 1-->take back money from mom

Mom gives back you $25

Now you have $25

Step 2: return the money you owe to friend

you owe $16

You have $25 and return $16

Thus , money you will have = money you have - money you return

money you have = $25 - $16 = $9

Thus, you will have $9 , after all is settled.

7 0
3 years ago
HELP PLSS* i need an answer and please no guessing
Amiraneli [1.4K]
7 7/8 in mixed number form.
7 0
3 years ago
Read 2 more answers
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
1 3/8 + 4 1/6 jwbhcjhbnegggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
Ilya [14]
BOIII get outta here
3 0
3 years ago
Read 2 more answers
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