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2 years ago

Lesson Posttes

Mathematics

1 answer:

joja [24]2 years ago

7 0

A translation 2 units to the left followed by dilation by 1/2.

The correct option is (B).

<h3>What is congruency?</h3>

It states that that two triangles are said to be congruent if they are copies of each other and when superposed, they cover each other exactly.

The complete question is:

Which of the following composition of transformations would create an

image that is not congruent to its original image?

A rotation of 45° followed by a reflection across the x-axis

A translation 2 units to the left followed by dilation by 1/2

A reflection across the y-axis followed by a rotation of 60°

A rotation of 135º followed by a translation of 4 units to the right.

After a translation of 2 units to the left followed by dilation by 1/2 the image will not be congruent to its original image.

All other options are reflection and rotation, which not fit into the situation.

Learn more about congruency here:

brainly.com/question/7888063

#SPJ1

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(-9, 19), (10, 16) Find the slope of the line through each pair of points

dolphi86 [110]

Answer:

-3/19

Step-by-step explanation:

slope = rise/run = (y1 - y2) / (x1 - x2)

s = (19 - 16) / (-9 -10)

s = 3/-19 = -3/19

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3 years ago

3. The masses of two gorillas are given below. A female gorila has a mass of

loris [4]

Answer: 135,000g

Hope this helped :)

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3 years ago

What is the slope of function 3x-5y=17

Sliva [168]

Answer:

y=5/3x-3.4

Step-by-step explanation:

slope is 5/3

6 0

3 years ago

Read 2 more answers

B-2-11. Find the inverse Laplace transform of s + 1/s(s^2 + s +1)

Aleksandr-060686 [28]

Answer:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)$

Step-by-step explanation:

let's start by separating the fraction into two new smaller fractions

First, s(s^2+s+1) must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.

$\frac{s+1}{s(s^{2} + s +1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}$

Where A, B and C are unknown constants. The numerator of s is a constant A, because s is linear, the numerator of s^2+s+1 is a linear expression Bs+C because s^2+s+1 is a quadratic expression.

Multiply both sides by the complete denominator:

$[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]$

Simplify, reorganize and compare every coefficient both sides:

$s+1=A(s^2 + s +1)+(Bs+C)(s)\\\\s+1=As^{2}+As+A+Bs^{2}+Cs\\\\0s^{2}+1s^{1}+1s^{0}=(A+B)s^{2}+(A+C)s^{1}+As^{0}\\\\0=A+B\\1=A+C\\1=A$

Solving the system, we find A=1, B=-1, C=0. Now:

$\frac{s+1}{s(s^{2} + s +1)}=\frac{1}{s}+\frac{-1s+0}{s^{2}+s+1}=\frac{1}{s}-\frac{s}{s^{2}+s+1}$

Then, we can solve the inverse Laplace transform with simplified expressions:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{\frac{1}{s}\}-\mathcal{L}^{-1}\{\frac{s}{s^{2}+s+1}\}$

The first inverse Laplace transform has the formula:

$\mathcal{L}^{-1}\{\frac{A}{s}\}=A\\ \\\mathcal{L}^{-1}\{\frac{1}{s}\}=1\\$

For:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}$

We have the formulas:

$\mathcal{L}^{-1}\{\frac{s-a}{(s-a)^{2}+b^{2}}\}=e^{at}cos(bt)\\\\\mathcal{L}^{-1}\{\frac{b}{(s-a)^{2}+b^{2}}\}=e^{at}sin(bt)$

We have to factorize the denominator:

$-\frac{s}{s^{2}+s+1}=-\frac{s+1/2-1/2}{(s+1/2)^{2}+3/4}=-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}$

It means that:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}\}$

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\mathcal{L}^{-1}\{\frac{1/2}{(s+1/2)^{2}+3/4}\}\\\\\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\frac{1}{2} \mathcal{L}^{-1}\{\frac{1}{(s+1/2)^{2}+3/4}\}$

So a=-1/2 and b=(√3)/2. Then:

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}=e^{-\frac{t}{2}}[cos\frac{\sqrt{3}t }{2}]\\\\\\\frac{1}{2}[\frac{2}{\sqrt{3} } ]\mathcal{L}^{-1}\{\frac{\sqrt{3}/2 }{(s+1/2)^{2}+3/4}\}=\frac{1}{\sqrt{3} } e^{-\frac{t}{2}}[sin\frac{\sqrt{3}t }{2}]$

Finally:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)$

7 0

4 years ago

In his free time, Gary spends 11 hours per week on the Internet and 11 hours per week playing video games. If Gary has five hour

Tpy6a [65]

Answer:

62.8% OR 0.628

Step-by-step explanation:

11 hours playing video games

11 hours on the Internet

11 + 11 = 22

5 hours of free time each day

7 days in a week

5 x 7 = 35

22/35 hours are spent

That is 0.62857142857 or I just mathmatically rounded to 0.628

0.628 into a fraction is 62.8%

6 0

3 years ago