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Yakvenalex [24]
3 years ago
6

The universal set is U = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}. If A = {Monday, Tuesday,Wednesday, Th

ursday, Friday} and B = {Friday, Saturday, Sunday}, find the indicated set.
A intersect B '
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

{Monday, Tuesday, Wednesday, Thursday}

Step-by-step explanation:

First we need to get the complement of B i.e B'

B' are the set in the Universal set but that are not in B

B' = {Monday, Tuesday, Wednesday, Thursday}

A= {Monday, Tuesday, Wednesday, Thursday, Friday}

A intersect B ' are the days of the week that are common to both sets;

A intersect B ' = {Monday, Tuesday, Wednesday, Thursday}

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A cubical reservoir has capacity of 480000 l itters of liquid. If ifs length and breadth are 10m and 8m respectively, find the h
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Volume = length \times breadth \times height \\\\480000 = 10 \times 8 \times  height\\\\480000 = 80 \times height\\\\height = \frac{480000}{80} =  6000 \ m

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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

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3 years ago
Which expressions are equivalent to <br> 2r+(t+r)
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Answer:

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