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Vectors u and v are shown on the graph. Part A: Write u and v in component form. Show your work. Part B: Find u + v. Show your work. Part C: Find 5u - 2v. Show your work.​

Vectors u and v are shown on the graph Part A Write u and v in component form Show your work Part B Find u v Show your work Part C Find 5u 2v Show your work class=

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msm555

Answer:

A) [tex]\sf u = (4, 8) \quad v = (4, -7) [/tex]

B) [tex]\sf u + v = (8, 1) [/tex]

C) [tex]\sf 5u - 2v = (12, 54) [/tex]

Step-by-step Explanation:

Part A: Writing vectors [tex]\sf u [/tex] and [tex]\sf v [/tex] in component form based on the provided vector operations:

Given:

[tex]\sf u = (6, 2) - (2, -6) = (6 - 2, 2 - (-6)) = (4, 8) [/tex]

[tex]\sf v = (11, 1) - (7, 8) = (11 - 7, 1 - 8) = (4, -7) [/tex]

So,

[tex]\sf u = (4, 8) [/tex]

[tex]\sf v = (4, -7) [/tex]

[tex]\dotfill[/tex]

Part B: Finding [tex]\sf u + v [/tex]:

Given:

[tex]\sf u = (4, 8) [/tex]

[tex]\sf v = (4, -7) [/tex]

Adding corresponding components:

[tex]\sf u + v = (4 + 4, 8 + (-7)) = (8, 1) [/tex]

Therefore, [tex]\sf u + v = (8, 1) [/tex].

[tex]\dotfill[/tex]

Part C: Finding [tex]\sf 5u - 2v [/tex]:

Given:

[tex]\sf u = (4, 8) [/tex]

[tex]\sf v = (4, -7) [/tex]

Multiplying each component of [tex]\sf u [/tex] by 5 and each component of [tex]\sf v [/tex] by 2:

[tex]\sf 5u = 5 \cdot (4, 8) = (5 \cdot 4, 5 \cdot 8) = (20, 40) [/tex]

[tex]\sf 2v = 2 \cdot (4, -7) = (2 \cdot 4, 2 \cdot (-7)) = (8, -14) [/tex]

Now, subtract [tex]\sf 2v [/tex] from [tex]\sf 5u [/tex]:

[tex]\sf 5u - 2v = (20, 40) - (8, -14) \\\\= (20 - 8, 40 - (-14)) \\\\= (12, 54) [/tex]

Therefore, [tex]\sf 5u - 2v = (12, 54) [/tex]

semsee45

Answer:

A)  u = < 4, 8 >    
      v = < 4, -7 >

B)  u + v = < 8, 1 >

C)  5u - 2v = < 12, 54 >

[tex]\hrulefill[/tex]

Step-by-step explanation:

[tex]\hrulefill[/tex]

Part A

The component form of a vector is given as < x, y >, where x signifies the horizontal displacement and y denotes the vertical displacement.

Given a vector with an initial point (x₁, y₁) and terminal point (x₂, y₂), the component form can be found by subtracting the coordinates of the initial point from the coordinates of the terminal point: < x₂ - x₁, y₂ - y₁ >.

Vector u:

  • Initial point (x₁, y₁) = (2, -6)
  • Terminal point (x₂, y₂)= (6, 2)

Therefore:

[tex]\mathbf{u} = \langle 6 - 2, 2 - (-6) \rangle\\\\\mathbf{u} = \langle 4, 8 \rangle[/tex]

Vector v:

  • Initial point (x₁, y₁) = (7, 8)
  • Terminal point (x₂, y₂) = (11, 1)

Therefore:

[tex]\mathbf{v} = \langle 11 - 7, 1 - 8 \rangle\\\\\mathbf{v} = \langle 4, -7 \rangle[/tex]

[tex]\hrulefill[/tex]

Part B

To find the sum of vectors u and v, we simply add their corresponding components:

[tex]\mathbf{u} + \mathbf{v} = \langle 4 + 4, 8 + (-7) \rangle\\\\\mathbf{u} + \mathbf{v} = \langle 8, 1 \rangle[/tex]

[tex]\hrulefill[/tex]

Part C

To find 5u  - 2v, we multiply each component of u by 5 and each component of v by 2, and then subtract the corresponding components:

[tex]5\mathbf{u} = 5 \times \langle 4, 8 \rangle = \langle20, 40 \rangle\\\\2\mathbf{v} = 2 \times \langle4, -7 \rangle = \langle 8, -14 \rangle[/tex]

Now, we subtract the scaled v from the scaled u:

[tex]5\mathbf{u} - 2\mathbf{v} = \langle 20 - 8, 40 - (-14) \rangle\\\\5\mathbf{u} - 2\mathbf{v} = \langle 12, 54 \rangle[/tex]

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