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Mkey [24]
4 years ago
11

Students on an academic team took a test to enter a math competition. They could take the test twice. Some students improved the

ir score, while others did not. The table shows the change in score for each student. Which student had the greatest overall change in his or her score?
Student Test Score Change
Joe –2
Nivea 3
Juanita 4
Li –3
Mathematics
1 answer:
katrin [286]4 years ago
6 0

Juanita had the largest change. (+4 points)

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t
Svet_ta [14]

Answer:

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}

[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}

[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]

Finally, the third vector of basis E is e3(t)=t^{2}

T[e3(t)]=T(t^{2})

in the equation : a2 = 1

T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}

Then

[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]

And that is the third column of [T]EE

Let's write our matrix

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

X=\left[\begin{array}{c}1&0&0\\\end{array}\right]

AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

Applying the coordinates 2,6 and -2 to the basis E we obtain

2+6t-2t^{2}

That was the original result of T[e1(t)]

8 0
3 years ago
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