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2 years ago

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Suppose T is the transformation from ℝ2 to ℝ2 that results from a reflection over the y-axis followed by a reflection over the x

-axis. Find the matrix A that induces T.Suppose T is the transformation from ℝ2 to ℝ2 that results from a reflection over the y-axis followed by a reflection over the x-axis. Find the matrix A that induces T.Suppose T is the transformation from ℝ2 to ℝ2 that results from a reflection over the y-axis followed by a reflection over the x-axis. Find the matrix A that induces T

1 answer:

LenKa [72]2 years ago

8 0

<u>Answer:</u>

A = \begin{pmatrix}-1 & 0 \\0 & -1\end{pmatrix}

<u>Step-by-step explanation:</u>

Let, there be a point (x , y) in R2

After reflection on Y axis, it will be ( -x , y) and thereafter if we reflect the image over X -axis , it will be (-x , -y)

So,

we need to find out a matrix

[tex]\begin{pmatrix}a & b \\c & d\end{pmatrix}[tex]

such that,

(x , y) [tex]\begin{pmatrix}a & b \\c & d\end{pmatrix}[tex] = (-x. -y)

So, we have the following set of equations.

ax + cy = -x ------------(1)

bx + dy= -y-----------(2)

We can get the solution (by inspection) as,

[tex]\begin{pmatrix}a & b \\c & d\end{pmatrix}[tex]

= [tex]\begin{pmatrix}-1 & 0 \\0 & -1\end{pmatrix}[tex]

So, the matrix is,

A = \begin{pmatrix}-1 & 0 \\0 & -1\end{pmatrix} (answer)

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Which expressions are equivalent to 7^{-2}\cdot 7^67

AleksandrR [38]

Answer:

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ 4}$

Step-by-step explanation:

The given expresion is;

${7}^{ - 2} \cdot {7}^{6}$

We apply the product rule of indices;

${a}^{m} \cdot \: {a}^{n} = {a}^{m + n}$

We apply this property to our expression to get:

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ - 2 + 6}$

Simplify the exponent on the right;

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ 4}$

We can't see the options, but I hope this helps!

3 0

2 years ago

Read 2 more answers

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$

Verification:

$560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10$

Both the conditions are satisfied. This, the criticism is invalid.

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2 years ago

Jen and Ariel are reading a 400 page novel for their literature class. Jen decides to read 100 pages the first day and 50 pages

Furkat [3]

Answer:

Option B is the correct answer.

Step-by-step explanation:

Jen decides to read 100 pages the first day and 50 pages each day thereafter.

Number of pages read by Jen per day = 50

Number of pages read by Jen on first day = 100

Ariel's progress on reading the book is represented by the linear function y = 40x + 80, where y is the total number of pages read after x days.

Number of pages read by Ariel per day = 40

Number of pages read by Ariel on first day = 40 + 80 = 120

Option A:

Jen reads 20 pages per day more than Ariel

Wrong

Option B:

Ariel reads 10 pages per day less than Jen.

Correct

Option C:

Ariel read 20 pages less the first day than Jen read.

Wrong

Option D:

The reading rate each day for Jen and Ariel is the same.

Wrong

Option B is the correct answer.

6 0

2 years ago

Read 2 more answers

Doug purchased land for $8,000 in 1995. The year value of the land depreciated by 4% each year thereafter. Use an exponential fu

Marta_Voda [28]

Answer:

Option C. $6,012

Step-by-step explanation:

we know that

The formula to calculate the depreciated value is equal to

$V=P(1-r)^{t}$

where

V is the the depreciated value

P is the original value

r is the rate of depreciation in decimal

t is Number of Time Periods

in this problem we have

t = 7 years

P = $8,000

r = 0.04

substitute in the formula above

$V = 8,000(1-0.04)^{7} = 6,012$

Hope this helps :)

4 0

2 years ago