. Trong không gian vector ℝ3 cho hai hệ vector
{ } 1 2 3 ( 2,2,5); (2,1,3); (3,1,3) U u m u u = − = = = , { } 1 2 3 ( ( 1,2,1); ,2,1); ( 1,7,4) V v m v m v = = = − −= .
a. Chứng tỏ U , V là hai cơ sở của ℝ3
.
b. Cho x m = − ( ,2, 3), tìm tọa độ của x theo cơ sở V .
c. Tìm PU V →
và PV U →
.
d. Sử dụng công thức đổi tọa độ, tìm tọa độ của x theo cơ sở U .

Which of the following values cannot be probabilities? 0.08, 5 divided by 3, startroot 2 endroot, negative 0.59, 1, 0, 1.44, 3 divided by 5 select all the values that cannot be probabilities. a. five thirds b. 1.44 c. 1 d. startroot 2 endroot e. three fifths f. 0.08 g. 0 h. negative 0.59

Asystem of equations has 1 solution. if 4x-y=5 is one of the equations , which could be the other equation ?

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.