Find the general expression for the slope of a line tangent to the c.pdf
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Find the general expression for the slope of a line tangent to the curve of y=3x^2+2x at the point P(x,y). Then find the slopes for x=-3 and x=1.5. Sketch the curve and tangent lines. I need mtan,slope for -3, slope for 1.5 and sketch. Solution slope - dy/dx = 6x+2 slope for x=-3 = 6(-3)+2 = -16 and slope for x=1.5 = 6(1.5)+2 = 11.
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Find the midpoint of the line segment joining the points P1 and P2..pdf
Find the midpoint of the line segment joining the points P1 and P2.
P1= (4,3)
P2= (-3,7)
Type and ordered pair. Use intergers or simplified fractions for any numbers in the expression.
Solution
Midpoint = ( (x1+x2)/2, (y1+y2)/2 )
= ( (4-3)/2, (3+7)/2 )
= (1/2, 5).
Find the mean of the random variable. Find the mean of the ran.pdf
Find the mean of the random variable.
Find the mean of the random variable. The random variable X is the number of people who
have a college degree in a randomly selected group of four adults from a particular town. Its
probability distribution is given in the table.
Solution
The mean is sum(x*P(x)).
Find the Laplace transform ofSolutionWe can rewrite the ab.pdf
Find the Laplace transform of
Solution
We can rewrite the above function like so
f(t) = (t2-8t+20)u(t-4)
f(t) = t2 u(t-4)- 8tu(t-4) + 20u(t-4)
Now we need to apply a trick and subtract and add 4
f(t) = (t-4+4)2u(t-4)- 8(t-4+4)u(t-4) + 20u(t-4)
f(t) = (t-4)2u(t-4)+8(t-4)u(t-4)+ 16u(t-4) - 8(t-4)u(t-4) -32u(t-4)+ 20u(t-4)
f(t) = (t-4)2u(t-4) + 4u(t-4)
Now we can take the laplace transform
F(s) = 2e-4s/s3 + 4e-4s/s.
Find the inverse Laplace Transform. Please show all work.Solutio.pdf
The document provides the solution to finding the inverse Laplace Transform of the function F(s) = e^-7s / ( s^2 + 2s + 1 + 1). By applying the inverse Laplace transform, the solution is found to be f (t) = e^(-t) sin t f(t+7), which shifts the original function f(t) by 7 units to the right.
Find the integral. Check your answer by differentiation.sin6 cos d.pdf
Find the integral. Check your answer by differentiation.sin6? cos ?d?
Solution
Let u = sin(t)
du = cos(t) * dt
So, int (sin^6tcostdt
becomes
integral (u^6 * du)
u^7/7 + C
(1/7)sin^7(t) + C ---> ANSWER
Check by deriving :
(1/7)*(7sin^6(t))*cos(t) + 0
sin^6(t)*cos(t) --> which was what we had originally
So, by check, we can safely say that our answer obtained was right.
Find the general form of the equation of the circle.Center at the .pdf
Find the general form of the equation of the circle.
Center at the point (-4,-3); containing the point (-3,3)
Solution
radius^2 = ( -4 - ( -3) )^2 + ( -3 -3 ) ^2
= 37
Eqn
( x +4) ^2 + ( y + 3)^2 = 37 (or) x^2 + y^2 + 8x + 6y = 12.
Find the indicated area under the standard normal curveTo the .pdf
Find the indicated area under the standard normal curve
To the right of z = - 1.41
You will need the standard normal tabel
The area to the right of z = -1.41 under the standard normal curve is (round to four decimal
places as needed)
Solution
Area = Pr(Z> -1.41) = 1 ? P(Z < -1.41) ? 1 ? 0.0681= 0.9319.
Find the identity element for the binary operator f Z × Z Z den.pdf
The document is a request for help finding the identity element of a binary operator f defined as f(x,y)=x+y-xy. The response provides the definition of an identity element e as the element such that f(x,e)=f(e,x)=x for all x. It is shown that for the given operator f, plugging in values for e yields e(1-x)=0, which is true for all x only when e=0. Therefore, the identity element is 0.
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Find the midpoint of the line segment joining the points P1 and P2..pdf
Find the midpoint of the line segment joining the points P1 and P2.
P1= (4,3)
P2= (-3,7)
Type and ordered pair. Use intergers or simplified fractions for any numbers in the expression.
Solution
Midpoint = ( (x1+x2)/2, (y1+y2)/2 )
= ( (4-3)/2, (3+7)/2 )
= (1/2, 5).
Find the mean of the random variable. Find the mean of the ran.pdf
Find the mean of the random variable.
Find the mean of the random variable. The random variable X is the number of people who
have a college degree in a randomly selected group of four adults from a particular town. Its
probability distribution is given in the table.
Solution
The mean is sum(x*P(x)).
Find the Laplace transform ofSolutionWe can rewrite the ab.pdf
Find the Laplace transform of
Solution
We can rewrite the above function like so
f(t) = (t2-8t+20)u(t-4)
f(t) = t2 u(t-4)- 8tu(t-4) + 20u(t-4)
Now we need to apply a trick and subtract and add 4
f(t) = (t-4+4)2u(t-4)- 8(t-4+4)u(t-4) + 20u(t-4)
f(t) = (t-4)2u(t-4)+8(t-4)u(t-4)+ 16u(t-4) - 8(t-4)u(t-4) -32u(t-4)+ 20u(t-4)
f(t) = (t-4)2u(t-4) + 4u(t-4)
Now we can take the laplace transform
F(s) = 2e-4s/s3 + 4e-4s/s.
Find the inverse Laplace Transform. Please show all work.Solutio.pdf
The document provides the solution to finding the inverse Laplace Transform of the function F(s) = e^-7s / ( s^2 + 2s + 1 + 1). By applying the inverse Laplace transform, the solution is found to be f (t) = e^(-t) sin t f(t+7), which shifts the original function f(t) by 7 units to the right.
Find the integral. Check your answer by differentiation.sin6 cos d.pdf
Find the integral. Check your answer by differentiation.sin6? cos ?d?
Solution
Let u = sin(t)
du = cos(t) * dt
So, int (sin^6tcostdt
becomes
integral (u^6 * du)
u^7/7 + C
(1/7)sin^7(t) + C ---> ANSWER
Check by deriving :
(1/7)*(7sin^6(t))*cos(t) + 0
sin^6(t)*cos(t) --> which was what we had originally
So, by check, we can safely say that our answer obtained was right.
Find the general form of the equation of the circle.Center at the .pdf
Find the general form of the equation of the circle.
Center at the point (-4,-3); containing the point (-3,3)
Solution
radius^2 = ( -4 - ( -3) )^2 + ( -3 -3 ) ^2
= 37
Eqn
( x +4) ^2 + ( y + 3)^2 = 37 (or) x^2 + y^2 + 8x + 6y = 12.
Find the indicated area under the standard normal curveTo the .pdf
Find the indicated area under the standard normal curve
To the right of z = - 1.41
You will need the standard normal tabel
The area to the right of z = -1.41 under the standard normal curve is (round to four decimal
places as needed)
Solution
Area = Pr(Z> -1.41) = 1 ? P(Z < -1.41) ? 1 ? 0.0681= 0.9319.
Find the identity element for the binary operator f Z × Z Z den.pdf
The document is a request for help finding the identity element of a binary operator f defined as f(x,y)=x+y-xy. The response provides the definition of an identity element e as the element such that f(x,e)=f(e,x)=x for all x. It is shown that for the given operator f, plugging in values for e yields e(1-x)=0, which is true for all x only when e=0. Therefore, the identity element is 0.
Find the indicated area under the standard normal curve. Between z=.pdf
Find the indicated area under the standard normal curve. Between z= -1.09 and z=1.09. The area
between z= -1.09 and z= 1.09 under the standard normal curve is? (Round to four decimal
places as needed)
Solution
Pr (z greater than -1.09 but smaller than 1.09)
1 - Pr (z is smaller than -1.09) - Pr (z is greater than 1.09)
= 1 - 0.1379 - 0.1379
=0.7242.
Find the identical value. 1. Z0.02Solution Type the keyboard .pdf
Find the identical value. 1. Z0.02
Solution
Type the keyboard shortcut ctrl+f or ctrl+h. This will bring up the find box and find and replace
box. there are options at the bottom of the window saying replace or replace all, find all etc. it
will give you a list of all the occurences of that number, word etc and you can click on them to
go to that cell within the worksheet..
Find the gradient of the given function. Assume the variables are re.pdf
Find the gradient of the given function. Assume the variables are restricted to a domain on which
the function is defined.
z = (x+y)e^(3y)
Solution
Then, f/x = e3y
and f/y = 3e3y(x+y) + e3y
Hence, grad f = if/x + jf/y = e3y(i + (3x+3y+1)j).
find the general solution in implicit form of the differential equat.pdf
find the general solution in implicit form of the differential equation csc y dx + sec^2 x dy = 0 by
separation of variables
Solution
dx/siny + dy/cos2(x) = 0
dx/siny =-dy/cos2(x)
dx cos2(x) = -dy siny
(1+cos2x)/2 dx = -siny dy
integrate on both sides
x/2 + sin(2x)/4 = cos(y) + C.
Find the followingAverage payment periodFixed assest turnover.pdf
Find the following:
Average payment period
Fixed assest turnover
Sales to working capital
Total assest turnover
Capital intensity
Debt ratio M Ch 3a Problems Dezto.mheducation.com/hm.tpx?--0.02562706850698704-
1455040771836 a Home | Chegg.com value 50.00 points Use the following financial statements
for Lake of Egypt Marina, Inc LAKE OF EGYPT MARINA, INC Balance Sheet as of December
31, 2012 and 2011 (in millions of dollars) 2012 2011 2012 2011 Assets Current assets Liabilities
and Equity Current liabilities Cash and marketable securities Accounts receivable Inventory $45
$36 32 148 40 223 Accrued wages and taxes Accounts payable Notes payable $40 20 24 32 35
30 Total Long term debt Stockholders\' equity Total $ 308 $ 216 $105 $ 76 $ 67 200 Fixed assets
$ 255 Gross plant and equipment Less: Depreciation 200 40 Preferred stock (4 million shares) $
4 16 Common stock and paid in surplus 16 (16 million shares) Retained earnings $167$ 160 308
Net plant and equipment Other long term assets 104 25 24 $ 192 184 $ 500 400 Total Total $
328 124 Total assets Total liabilities and equity $ 500 400 LAKE OF EGYPT MARINA, INC
Income Statement for Years Ending December 31, 2012 and 2011 (in millions of dollars) 2012
2011 Net sales (all credit) Less: Cost of goods sold $ 800 600 192 320 ross profits $ 480 $408 [D
11:03 AM 2/9/2016 Start Search the web and Windows Desktop
Solution
All values caluclated for 2012:
---------------------------------
Average payment period=(account payable*365)/COGS
=(35*365)/320= 39.92 days
Fixed assest turnover= sales/Fixed assets
=(800/192)=4.17
Sales to working capital= sales/(Working capital)
working capital= current assets-current liabilty
=800/(308-105)=3.94
Total assest turnover= sales/assets
=800/500=1.6
Capital intensity= assets/sales
=500/800=.625
Debt ratio= debt/Assets
=(500-328)/500=.344.
Find the followingBasic earnings power M Ch 3a Problems Dezto.mh.pdf
Find the following:
Basic earnings power M Ch 3a Problems Dezto.mheducation.com/hm.tpx?--
0.02562706850698704-1455040771836 a Home | Chegg.com value 50.00 points Use the
following financial statements for Lake of Egypt Marina, Inc LAKE OF EGYPT MARINA, INC
Balance Sheet as of December 31, 2012 and 2011 (in millions of dollars) 2012 2011 2012 2011
Assets Current assets Liabilities and Equity Current liabilities Cash and marketable securities
Accounts receivable Inventory $45 $36 32 148 40 223 Accrued wages and taxes Accounts
payable Notes payable $40 20 24 32 35 30 Total Long term debt Stockholders\' equity Total $
308 $ 216 $105 $ 76 $ 67 200 Fixed assets $ 255 Gross plant and equipment Less: Depreciation
200 40 Preferred stock (4 million shares) $ 4 16 Common stock and paid in surplus 16 (16
million shares) Retained earnings $167$ 160 308 Net plant and equipment Other long term assets
104 25 24 $ 192 184 $ 500 400 Total Total $ 328 124 Total assets Total liabilities and equity $
500 400 LAKE OF EGYPT MARINA, INC Income Statement for Years Ending December 31,
2012 and 2011 (in millions of dollars) 2012 2011 Net sales (all credit) Less: Cost of goods sold $
800 600 192 320 ross profits $ 480 $408 [D 11:03 AM 2/9/2016 Start Search the web and
Windows Desktop
Solution
Basic Earingns Power = Earning Before Interest & Taxes / Total Assets
------>> For 2012: $368 / $500 = 0.736 or 73.6%
------>> For 2011: $342 / $400 = 0.855 0r 85.5%.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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Lifelines of National Economy chapter for Class 10 STUDY MATERIAL PDF
The chapter Lifelines of National Economy in Class 10 Geography focuses on the various modes of transportation and communication that play a vital role in the economic development of a country. These lifelines are crucial for the movement of goods, services, and people, thereby connecting different regions and promoting economic activities.
BBR 2024 Summer Sessions Interview Training
Qualitative research interview training by Professor Katrina Pritchard and Dr Helen Williams
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Find the indicated area under the standard normal curve. Between z=.pdf
Find the indicated area under the standard normal curve. Between z= -1.09 and z=1.09. The area
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places as needed)
Solution
Pr (z greater than -1.09 but smaller than 1.09)
1 - Pr (z is smaller than -1.09) - Pr (z is greater than 1.09)
= 1 - 0.1379 - 0.1379
=0.7242.
Find the identical value. 1. Z0.02Solution Type the keyboard .pdf
Find the identical value. 1. Z0.02
Solution
Type the keyboard shortcut ctrl+f or ctrl+h. This will bring up the find box and find and replace
box. there are options at the bottom of the window saying replace or replace all, find all etc. it
will give you a list of all the occurences of that number, word etc and you can click on them to
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Find the gradient of the given function. Assume the variables are restricted to a domain on which
the function is defined.
z = (x+y)e^(3y)
Solution
Then, f/x = e3y
and f/y = 3e3y(x+y) + e3y
Hence, grad f = if/x + jf/y = e3y(i + (3x+3y+1)j).
find the general solution in implicit form of the differential equat.pdf
find the general solution in implicit form of the differential equation csc y dx + sec^2 x dy = 0 by
separation of variables
Solution
dx/siny + dy/cos2(x) = 0
dx/siny =-dy/cos2(x)
dx cos2(x) = -dy siny
(1+cos2x)/2 dx = -siny dy
integrate on both sides
x/2 + sin(2x)/4 = cos(y) + C.
Find the followingAverage payment periodFixed assest turnover.pdf
Find the following:
Average payment period
Fixed assest turnover
Sales to working capital
Total assest turnover
Capital intensity
Debt ratio M Ch 3a Problems Dezto.mheducation.com/hm.tpx?--0.02562706850698704-
1455040771836 a Home | Chegg.com value 50.00 points Use the following financial statements
for Lake of Egypt Marina, Inc LAKE OF EGYPT MARINA, INC Balance Sheet as of December
31, 2012 and 2011 (in millions of dollars) 2012 2011 2012 2011 Assets Current assets Liabilities
and Equity Current liabilities Cash and marketable securities Accounts receivable Inventory $45
$36 32 148 40 223 Accrued wages and taxes Accounts payable Notes payable $40 20 24 32 35
30 Total Long term debt Stockholders\' equity Total $ 308 $ 216 $105 $ 76 $ 67 200 Fixed assets
$ 255 Gross plant and equipment Less: Depreciation 200 40 Preferred stock (4 million shares) $
4 16 Common stock and paid in surplus 16 (16 million shares) Retained earnings $167$ 160 308
Net plant and equipment Other long term assets 104 25 24 $ 192 184 $ 500 400 Total Total $
328 124 Total assets Total liabilities and equity $ 500 400 LAKE OF EGYPT MARINA, INC
Income Statement for Years Ending December 31, 2012 and 2011 (in millions of dollars) 2012
2011 Net sales (all credit) Less: Cost of goods sold $ 800 600 192 320 ross profits $ 480 $408 [D
11:03 AM 2/9/2016 Start Search the web and Windows Desktop
Solution
All values caluclated for 2012:
---------------------------------
Average payment period=(account payable*365)/COGS
=(35*365)/320= 39.92 days
Fixed assest turnover= sales/Fixed assets
=(800/192)=4.17
Sales to working capital= sales/(Working capital)
working capital= current assets-current liabilty
=800/(308-105)=3.94
Total assest turnover= sales/assets
=800/500=1.6
Capital intensity= assets/sales
=500/800=.625
Debt ratio= debt/Assets
=(500-328)/500=.344.
Find the followingBasic earnings power M Ch 3a Problems Dezto.mh.pdf
Find the following:
Basic earnings power M Ch 3a Problems Dezto.mheducation.com/hm.tpx?--
0.02562706850698704-1455040771836 a Home | Chegg.com value 50.00 points Use the
following financial statements for Lake of Egypt Marina, Inc LAKE OF EGYPT MARINA, INC
Balance Sheet as of December 31, 2012 and 2011 (in millions of dollars) 2012 2011 2012 2011
Assets Current assets Liabilities and Equity Current liabilities Cash and marketable securities
Accounts receivable Inventory $45 $36 32 148 40 223 Accrued wages and taxes Accounts
payable Notes payable $40 20 24 32 35 30 Total Long term debt Stockholders\' equity Total $
308 $ 216 $105 $ 76 $ 67 200 Fixed assets $ 255 Gross plant and equipment Less: Depreciation
200 40 Preferred stock (4 million shares) $ 4 16 Common stock and paid in surplus 16 (16
million shares) Retained earnings $167$ 160 308 Net plant and equipment Other long term assets
104 25 24 $ 192 184 $ 500 400 Total Total $ 328 124 Total assets Total liabilities and equity $
500 400 LAKE OF EGYPT MARINA, INC Income Statement for Years Ending December 31,
2012 and 2011 (in millions of dollars) 2012 2011 Net sales (all credit) Less: Cost of goods sold $
800 600 192 320 ross profits $ 480 $408 [D 11:03 AM 2/9/2016 Start Search the web and
Windows Desktop
Solution
Basic Earingns Power = Earning Before Interest & Taxes / Total Assets
------>> For 2012: $368 / $500 = 0.736 or 73.6%
------>> For 2011: $342 / $400 = 0.855 0r 85.5%.
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LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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