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ludmilkaskok [199]
2 years ago
8

Prove: An odd number squared is an odd number. = (2n + 1)2 ​

Mathematics
1 answer:
Mnenie [13.5K]2 years ago
4 0

Answer:

{(2n + 1)}^{2}  = 4 {n}^{2}  + 4n + 1 = 2(2 {n}^{2}  + 2n) + 1

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Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona
andrey2020 [161]

Answer:

a. v.v' = v₁v₂ -  v₁v₂ = 0 b.  (20, -21)/29 and  (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' =  (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j  

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

=  v₁v₂ -  v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0
2 years ago
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I hope this helps you




-36-7-9


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The greatest common factor of 60w^5y^3 and 78wy^2 is
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The answer would be 6w^5y^3
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Answer:

Step-by-step explanation:

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