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Nesterboy [21]
3 years ago
13

Represents the sets with a Venn diagram: U = {1,2,3,4,5,6,7,8,9,10, 11, 12} A = {1,2,3,4,6, 12} B = {1,2,4,8}

Mathematics
1 answer:
pychu [463]3 years ago
7 0

Answer:

The required figure is shown below

Step-by-step explanation:

Consider the provided sets:

U = {1,2,3,4,5,6,7,8,9,10,11,12}

A = {1,2,3,4,6, 12}

B = {1,2,4,8}

We need to draw the Venn diagram.

Draw a rectangle which represents set U.

Now, Make circles for set A and set B. Every circle should overlap.

Write the element of each set as shown in figure.

Place elements inside each circle which represents the elements of corresponding sets.

Place common element of set in the section in which all circles shapes overlap.

The required figure is shown below

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Step-by-step explanation:

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A dinner plate has a circumference of 106.76 cm. What is the area of the dinner plate? Use = 3.14.
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C=2pr,  r=c/(2p)

a=pr^2, using r found above we get:

a=p(c^2/(4p^2))

a=(c^2)/(4p), since c=106.76 and we approximate pi as 3.14

a=(106.76^2)/(4*3.14)

a=11612.2176/12.56 cm^2

a≈924.54 cm^2  (to nearest one-hundredth of a square cm)
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Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

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Answer:

(2,3)

Step-by-step explanation:

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