The periods during which the surface of elevation is changing linearly. Year, Y Elevation, E (ft) 1965 4193 1970 4196 1975 4199 1980 4199 1985 4206 1990 4203 1995 4200

Question

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Answer 1

Question

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

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Answer 2

Question

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

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Answer 3

Question

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

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Answer 4

Question

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

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Answer 5

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Answer 6

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Answer 7

Chapter 1.4, Problem 44E

To determine

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Answer 8

Expert Solution & Answer

Answer to Problem 44E

During the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Formula used:

The slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Calculation:

Consider the table that provides elevation of the surface of Great Salt Lake in Utah for different years.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

The horizontal axis that is x- axis, denote the year and the vertical axis that is y -axis denote the elevation.

Recall slope is m=y−y0x−x0 where (x,y) is an arbitrary point and (x0,y0) is the base point.

Slope between 1965 and 1970 is the slope between the points (1965,4193) and (1970,4196) is, here arbitrary point is (1970,4196) and base point is (1965,4193) .

Slope of the line is,

m=y−y0x−x0=4196−41931970−1965=35

Therefore, slope is 3 ft every 5 years.

Slope between 1970 and 1975 is the slope between the points (1970,4196) and (1975,4199) is, here arbitrary point is (1975,4199) and base point is (1970,4196) .

Slope of the line is,

m=y−y0x−x0=4199−41961975−1970=35

Therefore, slope is 3 ft every 5 years.

Slope between 1975 and 1980 is the slope between the points (1975,4199) and (1980,4199) is, here arbitrary point is (1980,4199) and base point is (1975,4199) .

Slope of the line is,

m=y−y0x−x0=4199−41991980−1975=05=0

Therefore, slope is 0. No change in surface of elevation.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1980 and 1985 is the slope between the points (1980,4199) and (1985,4206) is, here arbitrary point is (1985,4206) and base point is (1980,4199) .

Slope of the line is,

m=y−y0x−x0=4206−41991985−1980=75

Therefore, slope is 7 ft every 5 years.

Slope between 1985 and 1990 is the slope between the points (1985,4206) and (1990,4203) is, here arbitrary point is (1990,4203) and base point is (1985,4206) .

Slope of the line is,

m=y−y0x−x0=4203−42061990−1985=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Slope between 1990 and 1995 is the slope between the points (1995,4200) and (1990,4203) is, here arbitrary point is (1995,4200) and base point is (1990,4203) .

Slope of the line is,

m=y−y0x−x0=4200−42031995−1990=−35

Therefore, slope is decreasing by 3 ft every 5 years.

Years during which slope is same depicts that surface of elevation is changing linearly.

Thus, during the period 1965 to 1975 and 1985 to 1995, the surface of elevation is changing linearly.

Answer 12

Ch. 1.2 - Identify the variables and parameters in the...Ch. 1.2 - Prob. 2E Ch. 1.2 - Compute the values of the following functions at...Ch. 1.2 - Compute the values of the following functions at...Ch. 1.2 - Compute the values of the following functions at...Ch. 1.2 - Compute the values of the following functions at...Ch. 1.2 - Graph the given points and say which point does...Ch. 1.2 - Graph the given points and say which point does...Ch. 1.2 - Graph the given points and say which point does...Ch. 1.2 - Graph the given points and say which point does...

The periods during which the surface of elevation is changing linearly. Year, Y Elevation, E (ft) 1965 4193 1970 4196 1975 4199 1980 4199 1985 4206 1990 4203 1995 4200

Concept explainers

Videos

To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Answer to Problem 44E

Explanation of Solution

Want to see more full solutions like this?

Chapter 1 Solutions

The periods during which the surface of elevation is changing linearly. Year, Y Elevation, E (ft) 1965 4193 1970 4196 1975 4199 1980 4199 1985 4206 1990 4203 1995 4200

Concept explainers

Videos

To calculate: The periods during which the surface of elevation is changing linearly. Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200

Answer to Problem 44E

Explanation of Solution

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To calculate: The periods during which the surface of elevation is changing linearly.

Year, YElevation, E (ft)19654193197041961975419919804199198542061990420319954200