Find (a) the orthogonal projection of b onto Col A and (b) a least-squares solution of Axequalsb. Aequals[Start 3 By 2 Matrix 1st Row 1st Column negative 1 2nd Column 4 2nd Row 1st Column 1 2nd Column 8 3rd Row 1st Column negative 1 2nd Column 4 EndMatrix ], bequals[Start 3 By 1 Matrix 1st Row 1st Column 12 2nd Row 1st Column negative 4 3rd Row 1st Column 2 EndMatrix ]a. The orthogonal projection of b onto Col A is ModifyingAbove Bold b with caretequals nothing. (Simplify your answers.)

in geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. by extension, a line and a plane, or two planes, in three-dimensional euclidean space that do not share a point are said to be parallel.

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92

the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

the root or roots of a polynomial are the values of the variable that makes the polynomial equal to 0. we can find the roots of a polynomial making the expression equal to 0 and then, isolating the variable (it it has just one variable).

for example, given the following polynomial:

the root or zero of the polynomial, can be found by making the expression equal to 0, and then, isolating the variable, so, finding the root we have:

therefore, we have that the root of the polynomial is equal to 2.

we must remember that the number of roots that a polynomial can have, will depend on the degree of the polynomial, a first degree polynomial have will have one single root, a second degree polynomial (quadratic expression) may have two roots and so.

hence the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

have a nice day!