Given that the two swimmers competed and Ursula's speed is 60 m/min while Andre's speed is 48 m/min. The distance that the Ursula will catch up with Andre will be:
distance=(relative speed)×(time)
relative speed=60-48=12 m/min
the two swimmers met at a distance of:
12×1
=12 meters
4d I think it is right hope that helps
Point slope form of a line:

where

is your slope and

is a point on your line.
For a line with a slope of 3/5 that passes through (10, -2)...

Let's plug these values into our formula.

Subtracting by a negative is the same thing as adding (the double negatives cancel), so let's simplify that.

There's our equation in point-slope form. (keep in mind that you're not going to want to distribute that 3/5 or subtract the 2 or anything. You want to keep it similar to the formula.
Answer:
D: The graph shows a line of best fit because the points are plotted evenly above and below the line.
Step-by-step explanation:
Answer:
- h(x) = -2x^2 -x +3
- Reservoir A releases the same amount of water as Reservoir B over 1 week
Step-by-step explanation:
<u>Given</u>:
f(x) = x^2 -7x +5
g(x) = 3x^2 -6x +2
h(x) = f(x) -g(x)
<u>Find</u>:
h(x)
h(1)
<u>Solution</u>:
The expression for h(x) is found by evaluating its definition:
h(x) = (x^2 -7x +5) -(3x^2 -6x +2)
h(x) = x^2(1 -3) +x(-7 -(-6)) +(5 -2)
h(x) = -2x^2 -x +3
Then h(1) is found by substituting 1 for x:
h(1) = -2(1^2) -(1) +3 = -2 -1 +3
h(1) = 0 . . . . difference in release amounts after 1 week is 0
The true statements are ...
- h(x) = -2x^2 -x +3
- Reservoir A releases the same amount of water as Reservoir B over 1 week