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.
Answer:
1. A. 12a^2 + 8a + 4
2. C. 4xy+16x-4y^2-16y
3. C. 64q^6r^8s^4
4. A. -4x^4y^6
Step-by-step explanation:
48a^3 + 32a^2 + 16a / 4a = (48a^3) / 4a + (32a^2) / 4a + (16a) / 4a = 12a^2 + 8a + 4.
So, the answer is A. 12a^2 + 8a + 4.
(2x-2y)(2y+8) = 4xy - 4y^2 + 16x - 16y.
So, the answer is C. 4xy+16x-4y^2-16y.
(-8q^3 * r^4 * s^2)^2 = (-8)^2 * q^6 * r^8 * s^4 = 64q^6r^8s^4.
So, the answer is C. 64q^6r^8s^4.
-12x^8y^8 / 3x^4y^2 = (-12 / 3) * (x^(8 - 4)) * (y^(8 - 2)) = -4x^4y^6.
So, the answer is A. -4x^4y^6.
Hope this helps!
Answer:
0=0
infinitely many solutions
Step-by-step explanation:
plug x=2y+3 as x into 3x-6y=9...so
3(2y+3)-6y=9, which has our x=2y+3 plugged in because that is what x equals
if you solve its
6y+9-6y=9
0y=0
y=0
making it 0=0
you can't go further from this
5 is a prime factor of 155. Dividing 155 by 5, we get 31. 1*5*31 = 155.
Answer:
7 million
Step-by-step explanation:
If you use your finger and continue the graph you will get to 7.\.
