Find the sum of the first six terms of the geometric series 1 + 4 + 16 + ...

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

The adults of a certain type of insect have a mean length of 0.6 inch. The students in a science class measured 10 insects of th

mixer [17]

Answer:

Step-by-step explanation:

3

7 0

2 years ago

Read 2 more answers

Line a is perpendicular to B.<br> if the slope of line A is -3/4 what is the line of B

tatyana61 [14]

Step-by-step explanation:

line a is perpendicular to B.

if the slope of line A is -3/4 what is the line of B

3 0

2 years ago

The official run time of a movie is 140 minutes.Your friend who has seen the movie says its about 2.3 hours

Elza [17]

A.
1 hour = 60 minutes ;
2 hour s= 120 minutes;
140 - 120 = 20 minutes ;
20 = 60 ÷ 3 ;
20 minutes represents ( 1 / 3 ) hour;
Then, 140 minutes represents 2 1 / 3 hours ;
2 1 / 3 = 7 / 3 = 2.(3) ;
B.
2.(3) > 2.3 ;
your friend underestimate the length of the movie ;

3 0

3 years ago

Someone help me please:)

Kryger [21]

Answer:

a. The time to strike the ground is 5 seconds

b. The time for the ball to be over than 96 ft is 1 second

Step-by-step explanation:

a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero

∴ s(t) = 0 ⇒ $80t-16t^{2}=0$