A. (−3, 3)
3x – 4y = 21
3(-3) - 4(3) = 21
-21 = 21 >>>>> not equal

B. (−1, −6)
3(-1) - 4(-6) = 21
21 = 21 >>>>>>>>>>Equal

C. (7, 0)
3(7) - 4(0) = 21
21 = 21>>>>>>>>>>equal

D. (11, 3)
3(11) - 4(3) = 21
21 = 21 >>>>>>>>>equal

5 0

2 years ago

Read 2 more answers

A cake manufacturer boxes Swiss chocolate and German chocolate cakes at one site. A box of Swiss chocolate cake contains 0.55 lb

Kobotan [32]

Answer:

2,500 German chocolate cake boxes.

1,500 Swiss chocolate cake boxes.

Step-by-step explanation:

Let 'S' be the number of Swiss chocolate cakes boxed and 'G' the number of German cholocate cakes boxed. If all of the available ingredients are used:

$0.55*S +0.59*G = 2,300\\0.45*S+0.41*G = 1,700$

Solving the linear system above:

$(0.55 *S +0.59*G)- \frac{0.55}{0.45} (0.45*S+0.41*G)= 2,300-\frac{0.55}{0.45}*1,700*\\0.088888*G = 222.222\\G=2,500\\S =\frac{2,300-2,500*0.59}{0.55}\\S=1500$

2,500 German chocolate cake boxes and 1,500 Swiss chocolate cake boxes can be made each day.

6 0

2 years ago

If →u and →v are the vectors below, find the vector →w whose tail is at the point halfway from the tip of →v to the tip of →v−→u

S_A_V [24]

Û = (-1, -1, -1)
^v = (2, 3, -5)
^v - û = (2 + 1, 3 + 1, -5 + 1) = (3, 4, -4)
Half way from ^v to ^(v - u) = ((3 - 2)/2, (4 - 3)/2, (-4 + 5)/2) = (1/2, 1/2, 1/2)
Halfway from û to ^v = ((2 + 1)/2, (3 + 1)/2, (-5 + 1)/2) = (3/2, 2, -2)

The required vector ^w = ((3/2 - 1/2), (2 - 1/2), (-2 - 1/2)) = (1, 1/2, -5/2)

5 0

2 years ago