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Lostsunrise [7]
3 years ago
15

PLEASE HELP ME!! It’s due

Mathematics
1 answer:
tatyana61 [14]3 years ago
5 0

Answer:

dont cheat

Step-by-step explanation:

cheaters never win bro

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A person is selected at random. If there are seven days in a week, what is the probability that the person was not born on a Tue
levacccp [35]

Answer:

5 out of 7.

Step-by-step explanation:

There are seven days in one week. They are asking what is the probability of the person chosen not being born in a Tuesday or Wednesday. There are five other days, so there are five other options.

I hope I helped you!

7 0
3 years ago
Madeline uses 58 beads to make a necklace she plans to sell 36 necklaces at each craft fair she goes to the summer she plans to
g100num [7]
<h3>♫ :::::::::::::::::::::::::::::: // Hello There ! //  :::::::::::::::::::::::::::::: ♫</h3>

➷  Multiply the number of beads per necklace by the number of necklaces:

58 x 36 = 2088

This is the number of beads needed per each craft fair.

Multiply this value by 4 to get the final value:

2088 x 4 = 8352.

She will need 8352 beads.

<h3><u>❄️</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

5 0
3 years ago
Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the dista
Kamila [148]

40mph for Car B I think. 45mph for Car A. 45(mph) times 4(h) is 180(distance) and 40 times 4.5 is also 180. Sorry if I'm wrong.

8 0
3 years ago
. Parallelogram ABCD,
anastassius [24]
Diagram not attached, please add the question.
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
3 years ago
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