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 =)
It wants it to be in slope-intercept form.
y=mx+b
We have to first find the slope and plug it into point-slope form.
y-y1=m(x-x1)
Find the slope of the second line. (I did this one first on accident)
Rise/run= 3/1= 3 The slope is 3. Plug that in along with the point (0,3)
y-3=3(x-0)
y-3=3x
Add 3 to the other side.
y= 3x +3 <- <em>for the second line</em><em>
</em>
Now, the second.
rise/run= 1/2= .5 Use point (6,0)
y-0=.5(x-6)
y= .5x-3
y=.5x-3 <- for the first line
I hope this helps!
~kaiker
7 cm: 35000 cm (7*5000)
2.8 cm: 14000 cm (2.8*5000)
2 cm: 10000 cm (2*5000)
F(x)=5x-2
y=5x-2
y=5(3)-2
y=15-2
y=13
so the answer is B. 13
