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faust18 [17]
1 year ago
11

Directions: Construct the bisector of the following figures.Note: please draw out each answer. I have a deadline due today. Than

k you.

Mathematics
1 answer:
IceJOKER [234]1 year ago
7 0

1)

The bisector is the midpoint of the curved line. To find it, place the compass at point A. Open it and draw an arc on both sides.

Without altering the compass, place it at point B and draw arcs on both sides to intersect the previous arcs.

Mark the two points of intersection. Draw a line touching both points. The point where the line touches line AB is the bisector

The sample is shown below

Point C is the bisector

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AleksAgata [21]

Answer:

99

Step-by-step explanation:

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3 years ago
A circle has a circumference of 907.46907.46 units. What is the diameter of the circle?
astra-53 [7]

question is wrong sir

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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

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Integrate both sides of

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

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\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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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}

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Answer:

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Step-by-step explanation:

6 0
3 years ago
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Step-by-step explanation:

Step 1

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and the number of adults in the group  be represented as  as y

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Step 2-- Solving

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