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Rina8888 [55]
3 years ago
7

Which one is the scaled copy Explain how

Mathematics
1 answer:
Kipish [7]3 years ago
7 0
C is the correct answer
C is the only option in which the original shaped can be multiplied and get another one.
explanation:
look at the dimensions of the original shape, 1x5x3
look at the dimensions of C,
2x10x6
multiply the original by 2 and the answer is C it is the only one that is consistant
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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

\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
Which graph represents an exponential function?
nadya68 [22]

Answer:

(You never showed the graphs we were suppose to choose from)

The graphs of all exponential functions have these characteristics. They all contain the point (0, 1), because a0 = 1. The x-axis is always an asymptote. They are decreasing if 0 < a < 1, and increasing if 1 < a.

6 0
2 years ago
Factor the Expression completely<br> -16x^2-6x^4
denpristay [2]

Answer: -2x^2 (3x^2 + 8)

Step-by-step explanation: a^b+c =a^b a^c

1) -16x^2 - 6x^ x^2

2) Rewrite -16 as -2 * 8

2.a Rewrite -6 as -2 * 3

= -2 * 8x^2 - 2 * 3x^2x^2

3) facter out common term: -2x^2

= ur answer

7 0
2 years ago
In a game of rolling a die. If the number showing is even, you win $3, if the number showing is 1 you win $3 and if the number s
IceJOKER [234]

Answer:

(a)

\left|\begin{array}{c|c|c}---&---&---\\x&\$0&\$3\\---&---&---\\P(x)&\dfrac13&\dfrac23\\---&---&---\\\end{array}\right|

(b)$2

Step-by-step explanation:

In the given game of rolling a die. these are the possible winnings.

  • If the number showing is even(2, 4, or 6) or 1, you win $3.
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There are 6 sides in the die.

P($obtaining a 1,2,4 or 6)=\dfrac46=\dfrac23\\P($obtaining a 3 or 5)=\dfrac26=\dfrac13

(i)The probability distribution of x.

Let x be the amount won

Therefore:

Probability distribution of x.

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(ii) Expected amount of dollar won

Expected Amount

=\sum x_iP(x_i)\\=(0*\dfrac13)+(3*\dfrac23)\\=\$2\\

You would expect to win $2.

7 0
3 years ago
A school choir needs to make T-shirts for its 75 members and has set aside some money in their budget to pay for them. The membe
Katyanochek1 [597]
50 + 3(B) + 50 + 3(R) + 50 + 3(Y) would be the equation.
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