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3 years ago

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How do I get this answer

1 answer:

sveticcg [70]3 years ago

6 0

Answer:

Round it otherwise round it because i

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Can someone help me figure out the answer?

love history [14]

Answer:

8763

Step-by-step explanation:

Let x represent the number of students the college had last year. Then this year's enrollment is ...

... x - 3%·x = 8500

... x(1 - 0.03) = 8500 . . . . . collect terms

... x = 8500/0.97 ≈ 8762.89 . . . . divide by the coefficient of x

Enrollment last year was about 8763.

_____

Of course, you know 3% = 3/100 = 0.03.

5 0

3 years ago

Roman55 [17]

I hoped this helped or something

5 0

3 years ago

Read the following statement.

nekit [7.7K]

Bro... ahahaha the answer is D DDDDDDDDDDDDDDD peace man

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:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago

There are two pizzas Conor ate 1/4 Brandon ate 2/8 Tyler ate 3/4 and Audrey ate. 4/8 who ate the most of the two pizzas

saul85 [17]

Tyler. 3/4 is the highest fraction of the four options.

7 0

3 years ago