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lord [1]
2 years ago
6

Solve each equation | n/5| +1=2

Mathematics
2 answers:
erastova [34]2 years ago
4 0

Answer:

|n \div 5|  + 1 = 2 \\  |n \div 5|  = 1 \\ n = 5

allochka39001 [22]2 years ago
3 0

Answer:

|n|= 5

Step-by-step explanation:

|n/5|+1=2

|n/5|=2-1

|n/5|=1

|n|=5×1

|n|=5

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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

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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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