Let a=⟨3,3⟩ and b=⟨3,−3⟩. show that there are scalars s and t so that sa+tb=⟨12,−18⟩ you might want to sketch the vectors to get some intuition.

The  sa+tb = <12, -18> can be written as .

a = <3, 3>

b = <3, -3>

resultant = <12, -18>

Solving for x side,

solving for y side,

substituting the value of s,

substituting the value of t,

Hence, the sa+tb = <12, -18> can be written as .

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The scalars are s=-1 and t=5 such that sa+tb=<12,−18>

Step-by-step explanation:

Let a=<3,3> and b=<3,-3>, s and t are scalars such that sa+tb=<12,-18>

Let d be a scalar quantity and v=<a, b> be a vector quantity.

Now to multiply the vector v=<a, b> by the scalar quantity d, that is multiply each of its components by d as below:

d<a, b>=<ad,bd>

Given that sa+tb=s<3,3>+t<3,-3>

=<3s,3s>+<3t,-3t>

sa+tb=<3s+3t,3s-3t>

Now equating it to <12,-18> we get

3s+3t=12 and 3s-3t=-18

3(s+t)=12

s+t=4

s=4-t

Now substitute the value s=4-t in 3s-3t=-18 we get

3(4-t)-3t=-18

12-3t-3t=-18

-6t=-18-12

-6t=-30

Therefore t=5

And s=4-5=-1

Therefore s=-1 and t=5 such that sa+tb=<12,−18>

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: