Not only you're gonna divide 8 by -3 for some reason, but you also divide all terms by -3. For instance,
(-3y = -2x + 8) / -3
(-3/-3)y = (-2/-3)x + 8/-3
y = (2/3)x - 8/3
Hope this helps.
Answer:
The speeds of the cars is: 0.625 miles/minute
Step-by-step explanation:
We use systems of equations in two variables to solve this problem.
Recall that the definition of speed (v) is the quotient of the distance traveled divided the time it took :
. Notice as well that the speed of both cars is the same, but their times are different because they covered different distances. So if we find the distances they covered, we can easily find what their speed was.
Writing the velocity equation for car A (which reached its destination in 24 minutes) is:

Now we write a similar equation for car B which travels 5 miles further than car A and does it in 32 minutes:

Now we solve for
in this last equation and make the substitution in the equation for car A:

So this is the speed of both cars: 0.625 miles/minute
I think it 4194304 i THINK thats the answer
The distance between the abscissas (x-coordinates) of the points is equal to 13. That of the ordinates (y-coordinates) is equal to 10. Thus, the area is obtained by multiplying these values. The answer is equal to 130 units squared.
Answer:
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Step-by-step explanation:
Q17

For Q18 and Q19.
If ΔABC ≅ ΔSRT (congruent), then

Q20
