Ne given vectors u and v, find (a) 3u (b) 3u +2v (c) v-2u. ests (a)3u(type your answer in terms of i and j) nten ucce

a) u = (a)i+(b)j , so 3u = (3a)i + (3b)j.

b) u as in a), v = (c)i+(d)j, so 3u + 2v = (3a+2c)i + (3b+2d)j

c) v-2u = (c-2a)i + (d-2b)j

Step-by-step explanation:

Lets consider u = (a,b) and v = (c,d).

a) When a vector is multiplied by a constant, each element of the vector is multiplied by the constant.

So 3u = 3(a,b) = (3a, 3b) = (3a)i + (3b)j.

b) First, we multiply both vectors by their respective constants.

3u = 3(a,b) = (3a, 3b) = (3a)i + (3b)j.

2v = 2(c,d) = (2c, 2d) = (2c)i + (2d)j.

Then, we add. When computing and addition between vector, we add the elements that are in the same position, i.e. (u+v)(1) = u(1)+v(1)...

So 3u + 2v = (3a, 3b) + (2c, 2d) = (3a+2c, 3b+2d) = (3a+2c)i + (3b+2d)j

c)

v = (c,d)

-2u = (-2a, -2b)

v-2u = (c-2a, d-2b) = (c-2a)i + (d-2b)j

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation:

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11-1 > 1-1+a

-12 > a