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ArbitrLikvidat [17]
3 years ago
10

Is math real? I feel like it doesn't exist.

Mathematics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

it does also can you help me with my question it is on my page. it is due tomorrow and i need help

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Need answers to three questions in picture will give brainly
Trava [24]

Answer:

3. the answer is A3

Step-by-step explanation:

you do 3x-1 and you get 3

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3 years ago
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If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.
alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]

The inverse of the matrix is

\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]

So, the Inverse Transformation is given by

T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})

So, no option is correct. And the answer is

T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})

Learn more about linear transformations here-

brainly.com/question/13005179

#SPJ10

5 0
2 years ago
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ohaa [14]

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Step-by-step explanation:

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2 years ago
How do i solve this problem?
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(4*3)+-4=8, this is just how to put it in the equation
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hoa [83]

Answer:

purple shape= 1824 inches squared

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