(1 point) let v be a vector space over a field k, and let uj, u2, u3, u4 be vectors in v. suppose u4 is a linear combination of {u} , u2, u3}. select the best of the following statements (i. e., select the statement that is true and gives the most information). a. we only know that span{uj, u2, u3 } c span{u1, u2, u3, u4}. b. span{u} , u2, u3 } = span{u} , u2, u3, u4} . c. span{uj, u2, u3 } = span{uj, u2, u3, u4 } only when u4 is a scalar multiple of one of {u}, u2, u3 }. d. there is no obvious relationship between span{u} , u2, u3 } and span{ui, u2, u3, u4 } . e. none of the above

a) false; the points do not have to be in a straight line.

step-by-step explanation:

the points could be in the shape of a triangle

b is incorrect

c is incorrect because it is saying the immediately don't form one when they are capable of doing so

d is incorrect

i dont know

step-by-step explanation:

d.) yes, because the volume of the cone is greater than the volume of the sphere

step-by-step explanation:

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.