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Mathematics

1 answer:

aleksley [76]3 years ago

4 0

Answer:

See brainly.com/question/23882263 for A and B parts of the first question

$HF = \sqrt{EH^2+EF^2} = \sqrt{2^2+3^2} = \sqrt{13}$

$d = \sqrt{2^2+5^2} = \sqrt{4+25} = \sqrt{29}$

Diagonals are perpendicular. Side length of the rhombus:

$a = \sqrt{(8/2)^2+(6/2)^2} = \sqrt{4^2+3^2} = \sqrt{25} = 5$

Perimeter:

P = 4a = 4(5) = 20

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Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane

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