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labwork
2 months ago
12

There were 340,000 cattle placed on feed. Write an equivalent ratio that could be used to find how many of these cattle were bet

ween 700 and 799 pounds. How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?

Mathematics
2 answers:
tester [12.3K]2 months ago
6 0
2/5 = x/340,000

Multiply 2 by 340,000 to get 680,000

Then, 680,000 divided by 5 results in 136,000

Consequently, x = 136,000

This means 136,000 cows were between 700 and 799 pounds


Leona [12.6K]2 months ago
3 0

A total of 340,000 cattle were placed on feed

What portion of the 340,000 cattle placed on feed fell within the weight range of 700 to 799 pounds?

The fraction representing the total cattle weighing 700 - 799 pounds is 2/5

Let x denote the count of cattle within the 700 - 799 pound range

We can set up a proportion using this fraction

\frac{2}{5} = \frac{x}{340,000}

We cross-multiply to solve for x

340000 * 2 = 5x

680000 = 5x

Dividing both sides by 5 gives us

Thus, x= 136,000

So, there are 136,000 cattle in the weight bracket of 700 and 799 pounds

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If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in
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Answer:

The highest achievable volume for the box is 2000000 cubic meters.

Step-by-step explanation:

Below is an outline of the volume (V), measured in cubic centimeters, and surface area (A_{s}), measured in square centimeters, for a box featuring a square base:

A_{s} = l^{2}+h\cdot l (1)

V = l^{2}\cdot h (2)

Where:

l - The length of the base's side, in centimeters.

h - The height of the box, in centimeters.

Using (2), we isolate h in the formula:

h = \frac{V}{l^{2}}

Then, we substitute into (1) and simplify the outcome:

A_{s} = l^{2}+ \frac{V}{l}

A_{s}\cdot l = l^{3}+V

V = A_{s}\cdot l -l^{3} (3)

Next, we calculate the first and second derivatives of this expression:

V' = A_{s}-3\cdot l^{2} (4)

V'' = -6\cdot l (5)

If V' = 0 and A_{s} = 30000\,cm^{2}, then we find that the critical value for the base's side length is:

30000-3\cdot l^{2} = 0

3\cdot l^{2} = 30000

l = 100\,cm

Subsequently, we assess this outcome using the second derivative's expression:

V'' = -600

According to Second Derivative Test, this critical value signifies an absolute maximum. Consequently, the largest volume obtainable for the box is:

V = 30000\cdot l - l^{3}

V = 2000000\,cm^{3}

The highest achievable volume for the box is 2000000 cubic meters.

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1 month ago
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