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3 years ago

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The figure shows the construction steps to duplicate an angle. The steps are in an incorrect order. What is the correct order of

the steps to duplicate an angle?
A. ADFECB
B. DAFEBC
C. DFAECB
D. ADEFCB

1 answer:

Maru [420]3 years ago

6 0

The correct order of the steps to duplicate an angle is A. ADFECB

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Which relation is not a function

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Answer:

B

Step-by-step explanation:

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<span>All real numbers that are greater than –6 but less than 6 is written as -6 < x < 6</span>

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Is (a + b) and (a - b)? perfect square trinomials?<br> O True<br> O False

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3 years ago

Read 2 more answers

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-

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2 years ago

12x - 2y = -1<br>+ 4x + 6y= -4<br>

svp [43]

Answer:

x = -7/40 , y = -11/20

Step-by-step explanation:

Solve the following system:

{12 x - 2 y = -1 | (equation 1)

4 x + 6 y = -4 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{12 x - 2 y = -1 | (equation 1)

0 x+(20 y)/3 = (-11)/3 | (equation 2)

Multiply equation 2 by 3:

{12 x - 2 y = -1 | (equation 1)

0 x+20 y = -11 | (equation 2)

Divide equation 2 by 20:

{12 x - 2 y = -1 | (equation 1)

0 x+y = (-11)/20 | (equation 2)

Add 2 × (equation 2) to equation 1:

{12 x+0 y = (-21)/10 | (equation 1)

0 x+y = -11/20 | (equation 2)

Divide equation 1 by 12:

{x+0 y = (-7)/40 | (equation 1)

0 x+y = -11/20 | (equation 2)

Collect results:

Answer: {x = -7/40 , y = -11/20

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3 years ago