bookishfairy73
contestada

PLease help and best gets brainliest.
School is 2 miles from home along a straight road. The table shows your distance from home as you walk home at a constant rate.

Time(mins): 10-20-30
Distnce(mi)1.5-1-0.5

1)Is the relationship in the table
proportional?
2)Find your distance from school fro each time in the table
3)Write an equation representing the relationship between the distance from school and time walking.


Respuesta :

CmdrWinters
1) Yes, the relationship in the table is proportional. If, when you've been walking for 10 minutes, you are 1.5 miles away from home, and when you've been walking for 20 minutes, you are 1 mile away from home, and when you've been talking 30 minutes, you are 0.5 miles away from home, then we can see that there is a proportion that happens here. For every 10 minutes you walk, you get 0.5 miles closer to your home.

2) We know that you've been walking 10 minutes already at the start of this problem, and we know that you walk at a steady pace of 0.5 miles every 10 minutes, so we just need to add 0.5 miles to our starting point to get the distance from the school to home, which makes it 2 miles away.

3) An equation representing the distance between the distance from school and time walking could be something like this:

t = 20d

Where t is the amount of time it takes to get home (in this case, t = 40 minutes) and d is the distance you can walk in 10 minutes (in this case, 0.5 miles)

The equation is lame, but that's the best I could do :\
Hope that helped =)
calculista

Part 1) Is the relationship in the table proportional?

Let

y-------> your distance from home in miles

x-------> the time in minutes

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form [tex] y/x=k [/tex] or [tex] y=kx [/tex]

Let

[tex]A(10,1.5)\\B(20,1)[/tex]

Find the slope AB

the slope is equal to

[tex] m=\frac{(y2-y1)}{(x2-x1)}[/tex]

Substitute the values

[tex] m=\frac{(1-1.5)}{(20-10)}[/tex]

[tex] m=\frac{(-0.5)}{(10)}[/tex]

[tex] m=-0.05[/tex]

Find the equation of the line with m and the point A

[tex] y-y1=m(x-x1) [/tex]

[tex] y-1.5=-0.05*(x-10) [/tex]

[tex] y=-0.05x+0.5+1.5 [/tex]

[tex] y=-0.05x+2 [/tex]

therefore

The answer part 1) is

the relationship in the table is not proportional

Part 2) Find your distance from school fro each time in the table

for

[tex]x=10\ minutes\\y=1.5\ miles\\ distance\ from\ school= 2-1.5=0.5\ miles[/tex]

for

[tex]x=20\ minutes\\y=1\ miles\\ distance\ from\ school= 2-1=1\ miles[/tex]

for

[tex]x=30\ minutes\\y=0.5\ miles\\ distance\ from\ school= 2-0.5=1.5\ miles[/tex]

Part 3) Write an equation representing the relationship between the distance from school and time walking

Let

y-------> your distance from school in miles

x-------> the time in minutes

[tex] A(10,0.5)\\ B(20,1)[/tex]

Find the slope AB

the slope is equal to

[tex] m=\frac{(y2-y1)}{(x2-x1)}[/tex]

Substitute the values

[tex] m=\frac{(1-0.5)}{(20-10)}[/tex]

[tex] m=\frac{(0.5)}{(10)}[/tex]

[tex] m=0.05[/tex]

Find the equation of the line with m and the point A

[tex] y-y1=m(x-x1) [/tex]

[tex] y-0.5=0.05*(x-10) [/tex]

[tex] y=0.05x-0.5+0.5[/tex]

[tex] y=0.05x[/tex]

therefore

the answer part 3) is

[tex] y=0.05x[/tex]

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