Question

Would someone help me with this work!!

Answer 1

See below and attached graph

Explanation:

The equation of a line in slope-intercept form is

`y= mx + b`

where m is the slope and y the y-intercept

Slope, m, is given by

`m=(y_(2)-y_(1))/(x_(2)-x_(1))`

where
`(x_(1),y_(1))` and
`(x_(2),y_(2))` are any two points on the line

I will compute the equations for two of the lines in Design A. Your team will have to work out the other equations

Design A

Line passing through (57,278) and (35,168)

`\text{\ensuremath{m=(168-278)/(35-57)}}`

`m=(-110)/(-22) = 5`

So equation of the line passing through these points is

`\bold{y=5x+b}`

To find b plug in any of the point values. Let's choose (35,168)

`168 = 5(35) + b`

`168 = 175 + b`

`b = 168-175 = -7`

Equation of this line is
`y= 5x-7`

Domain
`\text{D=(-\ensuremath{\infty},\ensuremath{\infty})}`

Range
`R=(-\infty,\infty)`

2. Line passing through (35,168) and (123,168)

Slope =
`(168-168)/(123-35)=0`

Equation is y = 0x + b or y = b

Plug in value for y = 168 to get the equation as

y = 168

Domain is
`D=(-\ensuremath{\infty},\ensuremath{\infty})`

Range R=168

The attached graph shows the equations for each line in Design A as well as their domains and ranges

Hope that helps