A. The ratio of lynx to mountain lions to wolverines is 2:3:1.
Thus, there are 2 lynx in the ratio. If there were 6 lynx, we would have to multiply all of the numbers in the ratio by 3 (because 6/2 = 3) to keep the ratio in the same proportion.
Therefore, because there is 1 wolverine in the ratio, and 1 * 3 = 3, if there were 6 lynx, there would be 3 wolverines.
b. We can use the same ideas that we had in part a to help us in part b.
There are 3 mountain lions in the ratio, but there are 15 mountain lions in the problem. Thus, the multiplier is 5, because 15/3=5.
Therefore, because there are 2 lynx in the ratio, and 2*5 = 10, if there were 15 mountain lions, there would be 10 lynx.
c. There is one wolverine in the ratio, but there are 10 wolverines in the problem. Thus, the multiplier is 10, because 10/1 = 10
Therefore, because there are 3 mountain lions in the ratio, and 3 * 10 = 30, if there were 10 wolverines in the park, then there would be 30 mountain lions.
d. The total number of lynx, mountain lions, and wolverines is 30.
To find out how many of each animal there should be, we must make an equation using the ratio and the variable x.
2x + 3x + 1x = 30
This equation means that the total number of animals together is 30, which is true. Now let's simplify by combining like terms.
6x = 30
Finally, we can simplify by dividing both sides by the coefficient of x, or 6.
x = 5
Thus, going back to our original equation, we know that the amount of lynx is 2x, mountain lions is 3x, and wolverines is 1x.
Lynx = 2x = 2(5) = 10 lynx
Mountain Lion = 3x = 3(5) = 15 mountain lions
Wolverines = 1x = 1(5) = 5 wolverines
Hope this helps! :)
Answer:
a = length of the base = 2.172 m
b = width of the base = 1.357 m
c = height = 4.072 m
Step-by-step explanation:
Suppose we want to build a rectangular storage container with open top whose volume is 12 cubic meters. Assume that the cost of materials for the base is 12 dollars per square meter, and the cost of materials for the sides is 8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?
lets call a = length of the base
b = width of the base
c = height
V = a.b.c = 12
Area without the top:
Area = ab + 2bc + 2ac
Cost = 12ab + 8.2bc + 8.2ac
Cost = 12ab + 16bc + 16ac
height = 3.width
c = 3b
Cost = 12ab + 16b.3b + 16a.3b = 12ab + 48b² + 48ab = 48b² + 60ab
abc = 12 → ab.3b = 12 → 3ab² = 12 → ab² = 4 → a = 4/b²
Cost = 48b² + 60ab = 48b² + 60b.4/b² = 48b² + 240/b
C(b) = 48b² + 240/b
C'(b) = 96b - 240/b²
Minimum cost: C'(b) = 0
96b - 240/b² = 0
(96b³ - 240)/b² = 0
96b³ - 240 = 0
96b³ = 240
b³ = 240/96
b³ = 2.5
b = 1.357m
c = 3b = 3*1.357 = 4.072m
a = 4/b² = 2.172m
Answer:
First option y=3/5x
Step-by-step explanation:
Because if you look on the graph, at x=5, y=3
in the equation, plug in x=5, you cancel out denominator left with y=3
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Answer:
2x^3-11x^2+16x-3
Step-by-step explanation:
Multiply x to all the other factors in the second parenthesis
Multiply -3 to all the other factors in the second parenthesis
