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defon
3 years ago
7

Write the inequality that represents the possible values for x

Mathematics
1 answer:
UNO [17]3 years ago
3 0

Answe8 times 8 equal 69 and bla ba

Step-by-step explanation:

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Rewrite 6√2 in the simplest form
mylen [45]
6<span>√2 is already in it's simplest form as a radical.


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5 0
3 years ago
When solved for x, which inequality represents the number line?
ikadub [295]

Answer:

x<-3

The value of x is found in all the numbers that are less than - 3 or all the numbers from - 3 to negative infinity

Please note: - 3 is not included in those numbers

4 0
2 years ago
How many U.S. dollars have the same value as 1 euro?
Natalija [7]

Answer:

1.13 dollars is equivalent to 1 euro

Step-by-step explanation:


7 0
3 years ago
Read 2 more answers
Maggie has $40 to buy a new dress and hair accessory for her school dance the dress she wants cost $35.75 the cost of her hair a
jonny [76]

Answer:

Yes Maggie has enough money to purchase the dress and hair accessory.

Step-by-step explanation:

Dress = $35.75

Hair accessory = $2.80

Total amount of money = $40

Does her have enough money?

Total Money in pocket - (Dress cost + hair accessory)

 $40.00 - ($35.75 + 2.80)

  $40.00 - ($36.00 + $3.00)     Round to the nearest dollar

  $40.00 - $39.00 = $1.00

Yes

8 0
3 years ago
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

\vec\jmath\times\vec k=\vec i

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

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

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

which means \vec f is indeed conservative and we can find f.

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}

\implies g(x,z)=4\sin(xz^2)+h(z)

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

\implies h(z)=C

for some constant C. So

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

3 0
4 years ago
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