Answer:
Infinite Solutions
Step-by-step explanation:
Try this: Combine 6 and 9, obtaining 15.
Compare this sum, 15, to the length of the 3rd side: 15.
If placed end to end, the 2 sides with lengths 6 and 9 form a new line of length 15. We'd end up with this new line superimposed on the 3rd side (with length 15).
Think: Do 2 straight lines of length 15, superimposed on one another, form a triangle?
1/10 of the thread is left because a connected to b is the thread and minus the 1/5 is 1/10 since 1/3 +total thread to begin wit.
Answer:
the correct answer is D 175
The dimensions of the enclosure that is most economical to construct are; x = 14.22 ft and y = 22.5 ft
<h3>How to maximize area?</h3>
Let the length of the rectangular area be x feet
Let the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 320 square feet
y = 320/x -----(1)
Cost to fence the three sides = $6 per foot
Therefore cost to fence one length and two width of the rectangular area
= 6(x + 2y)
Similarly cost to fence the fourth side = $13 per foot
So, the cost of the remaining length = 13x
Total cost to fence = 6(x + 2y) + 13x
Cost (C) = 6(x + 2y) + 13x
C = 6x + 12y + 13x
C = 19x + 12y
From equation (1),
C = 19x + 12(320/x)
C' = 19 - 3840/x²
At C' = 0, we have;
19 - 3840/x² = 0
19 = 3840/x²
19x² = 3840
x² = 3840/19
x = √(3840/19)
x = 14.22 ft
y = 320/14.22
y = 22.5 ft
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