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
The answer is 3 if you add 2 into 1 you get 3
So for this you start by figuring out all the numbers that add up to equal 12.
(0,12)(1,11)(2,10)(3,9)(4,8)(5,7)(6,6)
Then you put them into the equation 3l+2s=33 l=the larger number and s=the smaller number
To save space I'm only going to do the numbers that are the answers.
(3,9) 3(9)+2(3)=33
27+6=33
33=33
TADA
Hello :
<span> y=-6(2.5-x)(x-5.5) = -6(2.5x -13.75 -x² +5.5x)
y = -6(-x²+8x -13.75)
y = 6x²-48x+82.5
note :
if f(x) = ax²+bx +c the vertex is the point : ( -b/2a ; f(-b/2a))
a=6 b=-48 c = 82.5 .......calculate
-b/2a = -(-48)/2(6)= 4
f(4) =6(4)²-48(4)+82.5 =96 - 192 +82.5 = -13.5</span>
Hide and swing
It's where u have to hide and whenever you get caught you have to swing on vines to get away XD
