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aleksandrvk [35]
2 years ago
9

1. the prices for 6 different jerseys are

Mathematics
1 answer:
Musya8 [376]2 years ago
5 0

Answer:

Step-by-step explanation:

8

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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
A bike path is 3 miles long. there are distance markers every path one fourth mile to the end of the path. which number line cor
Fynjy0 [20]

Given :

A bike path is 3 miles long. there are distance markers every path one fourth mile to the end of the path.

To Find :

Which number line correctly models this situation and the total number of distance markers.

Solution :

It is given that their are markers every 1/4 part of mile.

So, their are 4 markers per mile.

Numbers of markers in 3 miles long path is :

n = 3\times 4\\\\n = 12

Therefore, the total number of distance markers in 3 mile path is 12.

Hence, this is the required solution.

5 0
3 years ago
Ashley deposits $500 into an account that earns 2% interest compounded 3 times per year. How much money will Ashley have in her
Vadim26 [7]
A=p (1+r/k)^kt
A=500×(1+0.02÷3)^(3×4)
A=541.50
5 0
3 years ago
Melinda is reading a story that has several characters. The character chloe is described in great detail and has several unique
mars1129 [50]

Answer:

round

Step-by-step explanation:

hope I helped

8 0
2 years ago
Combining like term 6k+7k
Nadya [2.5K]
13k is the answer :)
8 0
3 years ago
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