Use vector notation to describe the points that lie in the given configuration. (Let s and t be elements of the Reals.) The plane spanned by:

v1 = (6, 5, 0)
and
v2 = (0, 6, 5)
l(s, t) = ?

I(s,t) = {(6s, 5s+6t, 5t)| sER, tER }

Step-by-step explanation:

Given the data in the question;

A vector is in the span of V1 and V2 only if it is a linear combination.

of V1 and V2.

so 'V' is in the span if;

V = SV1 + tV2 [ S and t are real numbers ]

V =    [ 6si + 5sj ] + [ 6tj + 5tz]

V = 6s, 5s+6t, 5t

so

I(s,t) = {(6s, 5s+6t, 5t)| sER, tER }

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.

i got 10.28 but that rounds up to 10.3