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Lunna [17]
3 years ago
7

Simplify the following expression: 12a+9b2a-13a2b2+4ab2-16a

Mathematics
2 answers:
Rudiy273 years ago
8 0
12a+9b^2a-13a^2b^2+4ab^2-16a= \\ =12a-16a+9ab^2+4ab^2-13a^2b^2= \\ =-4a+13ab^2-13a^2b^2= \\ =a(-4+13b^2-13ab^2)
Elenna [48]3 years ago
6 0
12a + 9ab² - 13a²b² + 4ab² - 16a
12a - 16a + 9ab² + 4ab² - 13a²b²
-4a + 13ab² - 13a²b²
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3 years ago
16. Let W be the set of all vectors in R3 of the form a+2b b -3a Find a basis for Wand state the dimension of W.
Yuki888 [10]

Answer:

W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}

Observe that if the vector x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right] 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]

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]

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]\} is a basis for W and the dimension of W is 2.

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3 years ago
Find the slope of the line passing through the points (8,-3) and (4,-1)
docker41 [41]

\rm \to m =  \dfrac{y_2 - y_1}{x_2 - x_1}

\\  \\

\rm \to m =  \dfrac{ - 1 - 3}{4 - 8}

\\  \\

\rm \to m =  \dfrac{ - 4}{4 - 8}

\\  \\

\rm \to m =  \dfrac{ - 4}{ - 4}

\\  \\

\rm \to m = \cancel  \dfrac{ - 4}{ - 4}

\\  \\

\rm \to m = 1

7 0
2 years ago
Read 2 more answers
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