J)x^2 +2x +yx+2y-5x-10
x^2-3x+yx-10
k)5y^2(1-2y)
l)a(x+y+z)
Answer:
10 losses
Step-by-step explanation:
Here, we want to get the greatest possible number of games the team lost
Let the number of games won be x
Number drawn be y
Number lost be z
Mathematically;
x + y + z = 38
Let’s now work with the points
3(x) + 1(y) + z(0) = 80
3x + y = 80
So we have two equations here;
x + y + z = 80
3x + y = 80
The greatest possible number of games lost will minimize both the number of games won and the number of games drawn
We can have the following possible combinations of draws and wins;
26-2
25-5
24-8
23-11
22-14
21-17
21-17 is the highest possible to give a loss of zero
Subtracting each sum from 38, we have the following loses:
10, 8, 6, 4, 2 and 0
This shows the greatest possible number of games lost is 10
Answer:
76
Step-by-step explanation:
Normally you would do 76 * 0.25
For this I did 19 / 0.25
I hope this helps!
The bottom left one is the correct answer
