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3 years ago

Determine if the two figures are congruent and explain your answer using transformations.

Mathematics

1 answer:

Maslowich3 years ago

7 0

The coordinate A(-1, 1) reflected over y-axis is (1, 1) is the coordinate of E, hence the two figures are congruent

The given figures are quadrilaterals, in order to determine whether they are similar, we need to check if they are reflections of each other.

For the Quadrilateral ABCD, the coordinate of A is at A(-1, 1) and for the Quadrilateral DEFG, the coordinate of E is at E(1, 1).

Note that if an object is reflected over the y-axis the transformation is (x, y)->(-x, y)

We need to check whether if we reflect the coordinate A over the y-axis we will get coordinate E

Since the coordinate A(-1, 1) reflected over y-axis is (1, 1) is the coordinate of E, hence the two figures are congruent

Learn more on reflections here:brainly.com/question/1908648

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What is the domain of F(x)

nydimaria [60]

The domain of F(x) is all real numbers.

This is due to the fact that the function has no undefined points or domain constraints.

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3 years ago

Work out 25% of £18

konstantin123 [22]

It’s £4.5

Convert 25 to a decimal so 0.25

0.25 x 18 = 4.5

8 0

3 years ago

Read 2 more answers

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0

1 year ago

Can anyone explain me?

Alika [10]

Answer:

Step-by-step explanation:

a) (a + b)² = (a + b) * (a +b)

(a + b)³ = (a + b) * (a +b) * (a +b)

a²- b² = (a +b) (a - b)

Here (a + b) is common in all the three expressions

HCF = (a + b)

b) (x - 1) = (x - 1)

x² - 1 = (x - 1) * (x + 1)

(x³ - 1) = (x - 1) (x² + x + 1)

HCF = (x -1)

7 0

3 years ago

What is the slope of the line that passes through the points ( 5 , − 10 ) (5,−10) and ( 11 , − 12 ) ? (11,−12)? Write your answe

yan [13]

The slope of the line that passes through (5,−10) and ( 11 , − 12 ), in simplest form is: -1/3

Recall:

The slope of a line passing across through two points can be calculated using: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Given two points:

(5,−10) and ( 11 , − 12 )

Let,

$(5,-10) = (x_1, y_1)\\\\( 11 , - 12) = (x_2, y_2)$

Substitute

$m = \frac{-12 - (-10)}{11 - 5}$

Slope (m) = -2/6

Slope (m) = -1/3

Therefore the slope of the line that passes through (5,−10) and ( 11 , − 12 ), in simplest form is: -1/3

Learn more about slope of a line on:

brainly.com/question/3493733

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3 years ago