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gtnhenbr [62]
3 years ago
7

Asriel says 2 times 2 is 4 Adrian says 2 times 3 is 6 Brittne says 2 times 5 is 4.Who is correct?​

Mathematics
2 answers:
lys-0071 [83]3 years ago
5 0

Answer: Asriel & Adrian

Step-by-step explanation: 2x2 =4 (Ex: if you have 2 apples and was given 2 more you would have 4 in total)

2x3=6 (Ex: like if you have 2 oranges and was given 3 more you always have to count by your multiples.)

amid [387]3 years ago
4 0

Answer:

Asriel is correct because 2 times 2 is 4 and 2 times 3 is 6 and 2 times 5 is 10 so Asriel is correct.

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Evaluate the expression when a=4, b=−5, and c=−8. (−b)+c=
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Answer:

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Step-by-step explanation:

(−b)+c=

Let b = -5 and c = -8

- (-5) + -8

5 -8

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Consider the following two ordered bases of R3:
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Answer:

Let A = (a_1, ..., a_n) and B = (b_1, ..., b_n) bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors b_1, ..., b_n at base A.

The, we case correspond to find the coordinates of vectors of C,

\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right]   \}

at base B.

1. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]

Using rows operation we obtain the echelon form of the matrix

\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]

now we use backward substitution

c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4

Then the coordinate vector of \left[\begin{array}{ccc}2\\-1\\-1\end{array}\right] is \left[\begin{array}{ccc}4\\2\\1\end{array}\right]

2. We need to find a,b,c\in\mathbb{R} such that

\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

Then we find these values solving the linear system

\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]

Using rows operation we obtain the echelon form of the matrix

\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]

now we use backward substitutionc=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4

Then the coordinate vector of \left[\begin{array}{ccc}2\\0\\-1\end{array}\right] is \left[\begin{array}{ccc}4\\3\\2\end{array}\right]

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\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]

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\left[\begin{array}{cccc}1&-2&2&-3\\-1&2&-1&1\\0&-1&1&2\end{array}\right]

Using rows operation we obtain the echelon form of the matrix

\left[\begin{array}{cccc}1&-2&2&-3\\0&-1&1&2\\0&0&1&-2\end{array}\right]

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Then the coordinate vector of \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] is \left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right]

Then the change of basis matrix from B to C is

\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right]

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