Graph both functions to find the solution(s) to the system.
1 answer:
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Work the information to set inequalities that represent each condition or restriction.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
<span />
4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
<span />
4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.
The perimeter of a square is the sum of its sides and they are all equal, so to obtain the length of each of them we divide the perimeter of the first fence between 4:
P1= 64 feet/4 sides
P1= 16 feet
Then, the length of each side of the second fence will increase 2 feet at each end, as shown in the figure. We have then that the perimeter of the second fence is:
P2 = 20 feet x 4 sides
P2 = 80 feet
The sum of the perimeters of both fences is:
PT = P1 + P2
PT = 64 feet + 80 feet
PT = 144 feet
Total cost = 1.17 $ x 144 feet
Total cost = 168.48 $
The total cost of the fences was $ 168.48
