Answer:
yes
Step-by-step explanation:
For the relation to be a function, there must only be one corresponding value of y (range) for any value of x in the domain, that is a one-to- one relationship
This graph displays this feature and is a function.
Answer:
Step-by-step explanation:
L=3S
In 5 years,
L+5=2(S+5)
L+5=2S+10
L=2S+5
Substituting from above,
3S=2S+5
S=5
So then,
L=3(5)
L=15
Answer:
Distance =√(x₁ - y₁)²+ (x₂ - y₂)² = √97 = 9.85
Step-by-step explanation:
The Matrix X and Y could also be referred to as vectors in Rⁿ dimensions.
if Vector X = ( x₁ , x₂) and Vector Y = (y₁ , y₂)
then, Distance (X-Y) = ||X-Y|| = √(x₁ - y₁)²+ (x₂ - y₂)²
where, x₁ = 8, x₂ = -5 and y₁ = -1 , y₂ = -9
Distance = √(8 - (-1))²+ (-5 - (-9))² = √9² + 4² =√97 = 9.85
Answer:
![W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}](https://tex.z-dn.net/?f=W%3D%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3A%20a%2Cb%5Cin%5Cmathbb%7BR%7D%20%5C%7D)
Observe that if the vector ![x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=x%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
is in W then it satisfies:
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
This means that each vector in W can be expressed as a linear combination of the vectors ![\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Also we can see that those vectors are linear independent. Then the set
![\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%7D)
is a basis for W and the dimension of W is 2.
Step-by-step explanation:
solution.
Let S represent side of the equilateral triangle.
perimeter of equilateral ∆ =3×S
Therefore,if we let width to be represented by W
Width of rectangle=W
Length of rectangle=2W
one side of equilateral triangle= W+8
Therefore after analysing the question, the true statement is; The length of the rectangle is 2W
