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Verdich [7]
3 years ago
13

4 (2r + 3 ) = 15r - 11 +1 - 7r

Mathematics
2 answers:
KiRa [710]3 years ago
7 0

delete this

boop uwu

marissa [1.9K]3 years ago
3 0

Answer:

<em>No solutions</em>

Step-by-step explanation:

4(2r + 3) = 15r - 11 + 1 - 7r

Distribute:

8r + 12 = 15r - 11 + 1 - 7r

Combine like terms:

8r + 12 = 8r - 10

Same to both sides:

0r = -22

<em>No solutions</em>

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In a recent semester at a local university, 490 students enrolled in both General Chemistry and Calculus I. Of these students, 6
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Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -
irina [24]

Answer:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Step-by-step explanation:

Given

T:R^3->R^2

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

Required

Find the standard matrix

The standard matrix (A) is given by

Ax = T(x)

Where

T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

Ax = T(x) becomes

Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]

The x on both sides cancel out; and, we're left with:

A = [T(e_1)\ T(e_2)\ T(e_3)]

Recall that:

T(e_1) = (1,2)

T(e_2) = (-4,6)

T(e_3) = (2,-6)

In matrix:

(a,b) is represented as: \left[\begin{array}{c}a\\b\end{array}\right]

So:

T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]

T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]

T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]

Substitute the above expressions in A = [T(e_1)\ T(e_2)\ T(e_3)]

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

Hence, the standard of the matrix A is:

A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]

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