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ikadub [295]
3 years ago
10

How many mm are there in 3 m?​

Mathematics
2 answers:
kogti [31]3 years ago
7 0

Answer:

3000 mm

Step-by-step explanation:

3 m= 3×100= 300 cm

300 cm= 300×10= 3000 mm

hope it helps!!

Rina8888 [55]3 years ago
6 0

Answer:

3000 millimetres.....

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NemiM [27]
First you have to know BODMAS/BIDMAS (idk if all schools teach it like this) but this is the order in which you complete questions so it’s Brackets, Indices, Division/Multiplication, Addition/Subtraction
a) 5^2 is 5x5 which equals 25 1 to the power of anything is always 1 and 2^4 is 2x2x2x2 which is 16
So 5^2-1^5x2^4 = 25-1x16 which is then 25-16 = 9
b) 7^2+4^3 divided by 2^5 so 7^2 is 7x7 which = 49 then 4^3 means 4x4x4 = 64 and 2^5 is 2x2x2x2x2 which equals 32 so
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3 years ago
A cookie recipe calls for 1/4 cup of brown sugar for one batch of 24 cookies. How many cups of brown sugar would be needed to ba
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Answer:1 1/2

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3 years ago
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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

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tankabanditka [31]

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