Consider the list of n n-vectors:
1 1 1
0 1 1
a1= [ 0 ], a2= [ 0 ] ,, an= [ 1 ]
. . .
. . .
. . .
0 0 1

(The vector ai has its first i entries equal to one, and the remaining entries zero.)
Describe what happens when you run the Gram–Schmidt algorithm on this list of vectors, i. e., say what q1,..., qn are. Is a1,..., an a basis?

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right

77

step-by-step explanation: