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3 years ago

My brain doesn't feel like working today ;-;I just need the equation, not the answer

Mathematics

1 answer:

koban [17]3 years ago

4 0

Answer:

area = lengh x width is that enough of the formula? :D

Step-by-step explanation:

I'll give you the rest, just in case :P My brain got tired at the break before spring, here is the rest

400 = (L+30) X w

400 = 10+30 x 10 ( just b/c I can see 40 x 10 would be 400, calculus could also figure this out for us, but... that would be the minimum fencing he needs to buy and this questions isn't asking that, just make it fit on the back of his house, which we are not given, so just make up something, I used 10 b/c it was convenient )

400 = 40 x 10

2 sides are 40 , and one side is 10

so 90' of fence is needed

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4 years ago

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4 years ago

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storchak [24]

Answer:

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

Note that

$f(0) = (1,0,-1)$
$f'(\theta) = (\frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2}, cos(\frac{\theta}{\sqrt2})}, \frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2})$

Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.

The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.

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4 years ago

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