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2 years ago

Can someone help me find the missing x and the 2 Pythagorean’s triples

Mathematics

1 answer:

fgiga [73]2 years ago

4 0

Answer:

Below in BOLD.

Step-by-step explanation:

1. x^2 = 3^2 + 6^2

x^2 = 9 + 36 = 45

x = √45

x = 3√5.

2. 15^2 = x^2 + 12^2

x^2 = 15^2 - 12^2 = 225 - 144 = 81

x = 9.

9, 10 and 15 are Pythagorean triplets.

3. x^2 = 10^2 - 6^2 = 64

x = 8.

6, 8 and 10 are Pythagorean triplets.

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Answer:

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Step-by-step explanation:

The problem states that the box has square ends, so you can express volume with:

$v=x^{2} y$

Using the restriction stated in the problem to get another equation you can substitute in the one above:

$4x+y=84\\\\$

Substituting <em>y</em> whit this equation gives:

$v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}$

Now find the limit of <em>x</em>:

$\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14$

Find the length:

$y=84-4(14)=28$

You can now calculate the maximum volume:

$v=(14)^{2}(28)= 5488$

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Step-by-step explanation:

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Answer:

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