Let's start by assuming Armando's house is between Joey's and the park.
Let

be the distance Joey walked to Armando's house.
<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park.
</span>


That's probably the answer they're looking for. But what if the park is between Joey and Armando's houses or Joey is between the park and Armando? (The latter isn't really possible with the given distances.)
Let

be the distances between three collinear points like we have here. Our equation is really a few equations in one, something like

Let's get rid of the plus/minuses. Squaring,



For us, that's a quadratic equation for


I'll skip right to the solutions,


We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.
Answer:
Step-by-step explanation:
(4,6) and (20,14) are points on the line.
slope of line = (14-6)/(20-4) = 8/16= 1/2
point-slope equation for line of slope 1/2 that passes through (6,4):
y-4 = (½)(x-6)
in slope-intercept form:
y = ½x + 1
y-intercept = 1
Answer:
4pf
Step-by-step explanation:
Answer:
1/5
Step-by-step explanation:
Answer:
12 km/h
Step-by-step explanation:
Because they are asking for Pedro's average speed, you can divide the distance by the amount of time taken to travel. So in this case, Pedro's speed is equal to 96km/8h, which when divided simplifies to 12km/h. A little tip is to look at the units the answer is asking for (like in this problem it asks what was Pedro's average speed in km per hour). Km per hour is km/h, which will give you a hint on what units you will need to divide by. Hope this helps!