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miss Akunina [59]
3 years ago
15

Find the product. Tell which strategy you used.

Mathematics
1 answer:
Mrac [35]3 years ago
8 0

You need to provide a screenshot or recreation of the rest of the problem's information.

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Write the differential equation model that fits the given description. (a) The rate of change of the volume of a snowball (due t
miskamm [114]

Answer:

a) he rate of change of the volume of a snowball (due to melting) is proportional to the square of the volume at time t. Initially, the snowball has a volume of 900 cm3

\frac{dV(t)}{dT} = A*V(t)^{2} \\ and V(0) = 900cm^{3}

where A is a real constant, it appears because it says that the change i volume (dV/dt) is "proportional" to V(t)^{2}. Furthermore, we should assume that A is a negative number, because the volume of the snowball will decrease as the time pasese by.

(b) For an insect moving along some path, the velocity at time t is proportional to the square root of its position.

\frac{dr(t)}{dt}  = B*\sqrt{r(t)}

Here again appears a constant B for the "proportional" part. And i wrote the velocity as \frac{dr(t)}{dt} "the rate of change of the position with respect to te time".

7 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
This is an absolute value problem with LETTERS?! Absolute madness (no pun intended)
MAVERICK [17]

Answer:

a

Step-by-step explanation:

sjejeneissbebdejjesj

8 0
3 years ago
Upon registration, new members of the American cD club receive 30 cDs for review. Each month the club sends its members 2 more c
Pani-rosa [81]
520 CD's because of the 9 months and we add 30 time 9 we erase the 0 and multiply the 3 with the 9 it will be 27 right we add the zero because the originla number is 30
6 0
3 years ago
78 davided by 21 what is the anerw
ira [324]

Answer:

Step-by-step explanation:

3.714285714285714‬

3 0
3 years ago
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