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Tcecarenko [31]
3 years ago
5

What is the difference between a multiple and a factor?

Mathematics
2 answers:
Nina [5.8K]3 years ago
7 0
Factor are what we can multiply and multiples are what we get after multiplying , that's the difference between factors and multiples.
Inga [223]3 years ago
3 0
A multiple is like: 
Multiples of 5: 5, 10, 15, 20, 25
A factor is like:
Factors of 10:
1,2,5,10

Hope this helps!
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Solve for x, given the equation square root of x + 9 -4 = 1
saveliy_v [14]

Answer:

x = -4

Step-by-step explanation:

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3 years ago
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For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

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The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

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Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

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Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

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f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

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3 years ago
What's the answer ?????
Tanzania [10]
The awnser is a I hope I helped you out
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Maranda is buying pencils and a writing pad for her adult writing class. The writing pad she wants costs $3.50, and each pencil
Kruka [31]
Hello!

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Because both items would cost $3.50 together, Maranda will be eligible to buy them both and stay within her budget of $5.00.
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