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3 years ago

Describe in your own words how you can recognize the difference between exponential growth and exponential decay

1 answer:

7 0

Exponential growth and exponential decay can easily be recognized by directly observing the changes of a quantity may it be in charts or in diagrams. Exponential growth is the quick growth of quantity of a certain item in regular intervals while exponential decay is the reverse.

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Simplify 7N - 2N + 3P + 2P. (a) 5N + 5P (b) 10N + 10P (c)10NP

Yakvenalex [24]

7N - 2N + 3P + 2P

Combine like terms:
7N-2N=5N

3P+2P=5P

Put them together to get :
5N+5P

Final answer:
A

7 0

3 years ago

Read 2 more answers

Consider the equation 1+n+6/n+1=4/n-2

Inga [223]

A. The LCD is (n+1)(n-2)

B. n = -1, n = 2

C. n = negative 1 plus or minus the square root of 73, everything divided 2.

I’ve add a picture of the answers at the bottom in case I wasn’t clear on question C.

Hope that helps.

8 0

4 years ago

Read 2 more answers

A flat surface of a solid figure

-BARSIC- [3]

A flat surface of a solid figure is called a "face".
Your answer is face.

8 0

3 years ago

Read 2 more answers

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: