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2 years ago

Can anyone help with this math problem?

Mathematics

1 answer:

finlep [7]2 years ago

8 0

ok hi im here to help not rlly just want my points up your useless

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3 years ago

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3 years ago

Can you guys help please im so confused

marissa [1.9K]

Answer:

a) see below

b) 40x20 meters

Step-by-step explanation:

Write down what you know:

- The area of the enclosure is length*width, so

$A = x \cdot y$

- The length of the fencing is 80 meters, so

$2y + x = 80$

Now we have to combine these two equations above, and get rid of y in the process.

First rewrite the second as:

$y = 40 - \frac12 x$

Then substitute for y in the first:

$A = x (40 - \frac12 x) = 40x - \frac12x^2$

b) To maximize A, find the zero of the first derivative:

$A'(x) = 40 - x = 0 \implies x=40$

So y = (80-40)/2 = 20 meters.

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2 years ago

How to solve n-5*2/3=3/4

Scorpion4ik [409]

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3 years ago

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Answer:

1:0.06

Step-by-step explanation:

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