Well, this problem is best solved by setting up a system of two linear equations.
A linear equation can be defined
y=mx+b
where
b=initial value (when x=0), and
m=rate of increase or decrease.
In the given example, the x-axis represents hour, and the y-axis, number of cells.
Chemical #1
initial value = b = 12000 cells
rate = m = -4000 / hr
The equation is therefore
y1=-4000x+12000......................(1)
Similarly, for chemical #2
initial value = b = 6000 cells
rate = m = -3000 / hr
The equation is therefore
y2=-3000x+6000 .......................(2)
The time the two will have an equal sized colony would represent the solution of the system of equations (1) and (2), i.e. when y1=y2
which means
-4000x+12000 = -3000x+6000
transpose and solve for x
4000x-3000x = 12000-6000
1000x=6000
x=6 hours.
At 6 hours from the start,
y=-4000x+12000 = -4000*6+12000 = -24000+12000 = -12000 cells
So the solution is x=6, y=-12000, or (6,-12000)
Physical interpretation
Since cells cannot have a negative number, the two are actually equal before six hours, when they are both zero.
Case 1: y=0 when x=3
Case 2: y=0 when x=2
Therefore, after three hours, both trials will have zero cells.
You have to judge whether to give the mathematical solution (x=6,y=-12000) or the physical interpretation (x=3, y=0) as the answer.
Well this is were you would need to multiply the amount of pounds of rice they did and then divide the amount of months they give you!
Y = 8x - 2
y = 9x - 7
8x - 2 = 9x - 7
- 8x - 8x
-2 = x - 7
+ 7 + 7
5 = x
y = 8x -2
y = 8(5) - 2
y = 40 - 2
y = 38
(x, y) = (5, 38)
The answer is C.
Problem 7: Correct
Problem 8: Correct
Problem 9: Correct
The steps are below if you are curious
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Problem 7
S = 180*(n-2)
2340 = 180*(n-2)
2340/180 = n-2
13 = n-2
n-2 = 13
n = 13+2
n = 15
I'm using n in place of lowercase s, but the idea is the same. If anything, it is better to use n for the number of sides since S already stands for the sum of the interior angles. I'm not sure why your teacher decided to swap things like that.
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Problem 8
First find y
y+116 = 180
y+116-116 = 180-116
y = 64
which is then used to find x. The quadrilateral angles add up to 180*(n-2) = 180*(4-2) = 360 degrees
Add up the 4 angles, set the sum equal to 360, solve for x
x+y+125+72 = 360
x+64+125+72 = 360 ... substitution (plug in y = 64)
x+261 = 360
x+261-261 = 360-261
x = 99
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Problem 9
With any polygon, the sum of the exterior angles is always 360 degrees
The first two exterior angles add to 264. The missing exterior angle is x
x+264 = 360
x+264-264 = 360-264
x = 96
