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3 years ago

Round .0196 to thousandths

1 answer:

3 0

The 9 is in the thousandths place. The number after is it 6 (which it greater than 5) so 9 becomes a 10.

0.0196 rounded to the nearest thousandths is <u>0.02</u>

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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

Translation: 6 units right and 3 units down<br> s(-3, 3), C(-1, 4), W(-2,-1)

sdas [7]

S (3, 0)
C (5, 1)
W (4, -4)

Explanation
You take the first number and add 6 to it and you get the new number and then you take the second number and subtract 3 from it

S: -3 + 6 = 3
S- 3 - 3 = 0

C: -1 + 6 = 5
C: 4 - 3 = 1

W: -2 + 6 = 4
W: -1 - 3 = -4

3 0

2 years ago

What is the answer to this please

Alex_Xolod [135]

Domain: x≤5 (notice the closed dot on the right and the arrow on the left)
Range: y≤2 (notice the closed dot on the right and the arrow on the left)

The arrow means it keeps going infinitely. The closed dot means it stops there but is inclusive of that value.

3 0

3 years ago

Can someone please help me

Contact [7]

Answer:

the mid point is 6,3

Step-by-step explanation:

the mid point is 6,3

3 0

3 years ago

What is 53.67 to one decimal place

MAXImum [283]

53.67 to one decimal place is 53.7 since you round the .6 to the .7

8 0

2 years ago

Read 2 more answers