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3 years ago

If you had $20 US, estimate how many Yen and Pounds you would have. Show your work. please help

Mathematics

2 answers:

AlladinOne [14]3 years ago

5 0

I’m not trying to scam you I’ll do the work but can you give more details

GuDViN [60]3 years ago

3 0

Answer:

$20 USD equal 2094 JPY

Step-by-step explanation:

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Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your

love history [14]

If we substitute $x=r\cos\theta$ and $y=r\sin\theta$ , we get $r^2=x^2+y^2$ , so that

$z=\cos(x^2+y^2)=\cos(r^2)$

which is independent of $\theta$ , which in turn means the surface can be treated like a surface of revolution.

Consider the function $f(t)=\cos(t^2)$ defined over $0\le t\le1$ . Revolve the curve $C$ described by $f(t)$ about the line $t=0$ . The area of the surface obtained in this way is then

$\displaystyle2\pi\int_C\mathrm dS=2\pi\int_0^1\sqrt{1+f'(t)^2}\,\mathrm dt$

$=\displaystyle2\pi\int_0^1\sqrt{1+(-2t\sin(t^2))^2}\,\mathrm dt$

$=\displaystyle2\pi\int_0^1\sqrt{1+4t^2\sin^2(t^2)}\,\mathrm dt\approx7.4144$

4 0

3 years ago

PLS HELP ME ASAP!!!!

elena55 [62]

Answer:

2Pi/8

Step-by-step explanation:

6 0

3 years ago

The cylinder’s diameter and height are both 39.2 millimeters. What is the surface area of this cylinder?

Dominik [7]

Answer:

A = 7,241.2454026112 square mm.

or exact : A = 2,304.96‬*π sq mm

Step-by-step explanation:

Cylinder formula: A = π* r^2 + 2 *π * r* h + π*r^2

is the surface area.

d = 2r = 39.2 mm

h = 39.2 mm

r = 19.6 mm

A = π* r^2 + 2 *π * r* h + π*r^2

A = 2* π* (19.6)^2 + 2π (19.6)*(39.2)

A = 2,413.748467537 + 4,827.496935

A = 7,241.2454026112 square mm.

A = 2*pi* (768.32‬ + 384.16)

A = 2*1,152.48‬* pi

A = 2,304.96‬*π sq mm

8 0

3 years ago

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: