The width of the court is 30 feet!
We want to see how many solutions has an equation given some restrictions on the vectors of the equation.
We have 3 vectors in R2.
v₁, v₂, and v₃.
Where we know that v₁ and v₂ are parallel. And two vectors are parallel if one is a scalar times the other.
Then we can write:
v₂ = c*v₁
Where c is a real number.
Then our system:
x*v₁ + y*v₂ = v₃
Can be rewriten to:
x*v₁ + y*c*v₁ = v₃
(x + y*c)*v₁ = v₃
Assuming x, y, and c are real numbers, this only has a solution if v₁ is also parallel to v₃, because as you can see, the equation says that v₃ is a scallar times v₁.
Geometrically, this means that if we sum two parallel vectors, we will get a vector that is parallel to the two that we added.
If you want to learn more, you can read:
brainly.com/question/13322477
The correct option is: Option (B)

Explanation:
First thing is that the difference between each number in the series with the next number is 5. It means it must be the multiple of 5. There are two options that contain multiples of 5: Option B and Option D. Now in the option D, the upper limit is 6. If we put 6 in the expression: 5(6)-2, the last term would be 28. However in the series given in the question, the last term is 33. Hence 5(7) - 2 = 35 - 2 = 33 which is Option B.
When i=1: 5(1)-2 = 3
When i=2: 5(2)-2 = 8
When i=3: 5(3)-2 = 13
.
.
When i=7: 5(7)-2 = 35-2 = 33
Hence the correct option is (B).
Answer:
206
Step-by-step explanation:
70% of 294
70/100 × 294/1
205.8
round it off to 206
Answer:
£38,6267
Step-by-step explanation:
First year depreciation will be:
42800 * 0.(100-5) = £40,660
Second year worth after depreciation:
40660 * 0.(100-5) = £38,627
I hope I answered your question well.