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Maurinko [17]
3 years ago
13

What is the equation of the asymptote of the graph of y=2(6^x) ?

Mathematics
1 answer:
Arlecino [84]3 years ago
6 0
The value of can never be 0  


y = 0   



A
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-4

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Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

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8 0
3 years ago
Which logarithmic equation is equivalent to the exponential equation below?
anzhelika [568]

Answer:

log_e^{67.21}= a

Step-by-step explanation:

The equation is ,

\implies 67.21 = e^a

  • Here e is the Euler's Number . The log which are defined on the base e are called Natural logarithms.

Say if we have a expoteintial equation ,

\implies a^m = n

In logarithmic form it is ,

\implies log_a^n = m

Similarly our required answer will be ,

\implies\boxed{ log_e^{67.21}= a  }

3 0
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