Find the closest point to y in the subspace w spanned by bold v 1v1 and bold v 2v2. yequals=left bracket start 4 by 1 matrix 1st
row 1st column negative 15 2nd row 1st column negative 1 3rd row 1st column 1 4st row 1st column 2 endmatrix right bracket −15 −1 1 2 , bold v 1v1equals=left bracket start 4 by 1 matrix 1st row 1st column 1 2nd row 1st column 1 3rd row 1st column negative 1 4st row 1st column negative 2 endmatrix right bracket 1 1 −1 −2 , bold v 2v2equals=left bracket start 4 by 1 matrix 1st row 1st column 5 2nd row 1st column 1 3rd row 1st column 0 4st row 1st column 3 endmatrix right bracket
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Check the picture below.
notice, that one go-around is 2π, or namely 4π/2, and if we go 3π/2 more over, we'd end up at at 4π/2 + 3π/2 or 7π/2.
a couple of coterminal angles will be those in blue, one angle in the first go-around, and another one by simply going around one more time by 4π/2 extra, 7π/2 + 4π/2 is just that one there.
notice, that one go-around is 2π, or namely 4π/2, and if we go 3π/2 more over, we'd end up at at 4π/2 + 3π/2 or 7π/2.
a couple of coterminal angles will be those in blue, one angle in the first go-around, and another one by simply going around one more time by 4π/2 extra, 7π/2 + 4π/2 is just that one there.
Answer:
The answer is C. 3x+y=7, 6x+2y=10
Step-by-step explanation:
