Answer:
x=28
Step-by-step explanation:
vertical angles are congruent
3x-12=72
3x=84
x=28
Answer: total profit = $418
======================================================
Work Shown:
June
Income = (4 lawns)*($27 per lawn) = $108
Expenses = ($32 for gas)+($12 for trim line) = $44
Profit = income - expenses = 108-44 = $64
----------------------------
July
Income = (12 lawns)*($20 per lawn) = $240
Expenses = ($89 for gas)+($29 for blade sharpening) = $118
Profit = income - expenses = 240 - 89 = $151
----------------------------
August
Income = (16 lawns)*($20 per lawn) = $320
Expenses = ($101 for gas)+($16 for oil) = $117
Profit = income - expenses = 320-117 = $203
----------------------------
Total profit = (june profit)+(july profit)+(august profit)
Total profit = (64) + (151) + (203)
Total profit = $418
If the final result was negative, then we would call this a loss. However, we have a positive value, so we go with a profit.
Answer:
the ans is in the picture with the steps
(hope it helps can i plz have brainlist :D hehe)
Step-by-step explanation:
Answer:
correct answer is option B
Step-by-step explanation:
as we know equation of line
y = m x + c....................(1)
where c is the intercept
m is the slope
m = 
m = 
m = 35
putting value of m in equation (1)
we get
y = 35 x + c
from the graph we can clearly see that line is passing through (0,0) hence
0 = 35 (0)x + c
c = 0
hence the equation of line comes out to be
y = 35 x
correct answer is option B
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
