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Maurinko [17]
4 years ago
15

Sarah is turning 21 today. In the mail, she received 16 envelopes, and each envelope included $2.10. How much money did she rece

ive in the mail?
Mathematics
1 answer:
Mademuasel [1]4 years ago
7 0
Sarah will have $33.60 in total
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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
(40-3x7-9-6x2]+ 4+19<br> I need the answer to this
Schach [20]

-3x7=-21

6x2= -12

40+21-9+12

61-21

40+4+19

44+19= 63

5 0
3 years ago
Consider the function f(x)=x3+15x2+74x+120. If f(x)=0 for x=−6, for what other values of x is the function equal to 0? List the
jasenka [17]

Answer:

other x values are -5,-4

Step-by-step explanation:

f(x)=0 for x=−6

Apply synthetic division to get the quotient . using that we find other two values of x

-6                1           15             74          120

                   0          -6           - 54          -120                    

                  -------------------------------------------------

                    1           9             20         0

the quotient is x^2+9x+20=0

now factor the left hand side, product is 20 and sum is 9

x^2+9x+20=0\\(x+5)(x+4)=0\\x+5=0 , x=-5\\x+4=0, x=-4

So  other x values are -5,-4

8 0
3 years ago
Jasmine and Luke used fea part b
Nostrana [21]

Answer:

huh

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
Find the values of x and y
Mkey [24]

Answer:

x = 78º

y = 35º

Step-by-step explanation:

These are isosceles triangles

two angles are congruent

-----------------------------

x + 51 + 51 = 180

x = 180 - 102

x = 78º

y + y + 110 = 180

2y = 70

y = 35º

3 0
3 years ago
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