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>

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3