All vectors are in rn. check the true statements below:

a. for any scalar c, u* (cv) = c(u*v).

b. for a square matrix a, vectors in col a are orthogonal to vectors in nul a.

c. if vectors v1; : : : ; vp span a subspace w and if x is orthogonal to each vj for j = 1; : : : ; p, then x is in w(

d. if the distance from u to v is equal to the distance from u to -v, then u and v are orthogonal.

e. v*v =llvll^ 2

x/3-10=8

x/3=18

x=48

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

answer: the distance

step-by-step explanation:

to calculate the distance d of a line segment with endpoints (x1, y1) and (x2, y2) use the formula d (x2 x1)2 (y2 y1)2. to calculate the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) use the formula