Least squares and qr factorization. suppose a is an m n matrix with linearly independent columns and qr factorization a = qr, and b is an m-vector. the vector a^x is the linear combination of the columns of a that is closest to the vector b, i. e., it is the projection of b onto the set of linear combinations of the columns of a. (a) show that a^x = qqt b. (the matrix qqt is called the projection matrix.) (b) show that = bk2. (this is the square of the distance between b and the closest linear combination of the columns of a.)

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step-by-step explanation:

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

im not sure

step-by-step explanation: