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Alex17521 [72]
2 years ago
15

Jeff is responsible for buying supplies of food and medication for the dogs and cats at a local shelter. The food and medication

for a dog cost twice as much as for a cat. He needs to provide supplies for 164 cats and 24 dogs, and his budget is $4,240. How much can he spend on each dog for food and medication?
A. $50

B. $40

C. $30

D. $60

E. $45
Mathematics
1 answer:
NARA [144]2 years ago
4 0
He can spend $40 on each dog for food and medication.
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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)]

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What is the value of this expression?
jekas [21]

Answer:

108

Step-by-step explanation:

we first add 7+8=15 then 15×2=30 30+6=36 36×2=108

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A cube has a volume of 125 cubic inches. What is the length of each edge?
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We know the volume is 125 cubic inches.

125 = a^3

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5 = a

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