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nadya68 [22]
3 years ago
6

For ΔABC, which transformation composition is COMMUTATIVE?

Mathematics
2 answers:
liq [111]3 years ago
4 0
Commutative transformation is type of transformation in which the order of operations doesn't affect the result of transformation.
The only commutative transformation is D.
vladimir2022 [97]3 years ago
3 0

Answer:

<h3>option: D.</h3>

Step-by-step explanation:

We are asked  which transformation composition is COMMUTATIVE i.e. <em>we are asked to find that the resultant figure is independent of the operations applied to it  in any order.</em>

  • Also we know that translation and rotation together are not commutative hence, A and B options are incorrect.
  • Also rotation and reflection applied to a figure are not commutative.
  • But a pair of  reflection applied to a figure will be commutative

Hence, option: D is correct. (reflect across the y-axis and then reflect across the x-axis.)



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The square of the quantity three less than a number n is seven(answer in an algebraic expression)
ira [324]

Answer:

(n - 3)^2 = 7

Step-by-step explanation:

From the question, we understand that the number is n.

Three less than n is:

n - 3

The square is:

(n - 3)^2

The square equals 7;

So, we have:

(n - 3)^2 = 7

5 0
3 years ago
Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.
irina1246 [14]

Answer:

The solution of the diferential equation is:

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

Step-by-step explanation:

Given y" + y = te^{t}; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[te^{t}]

ℒ[y"]+ℒ[y]=ℒ[te^{t}]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[te^{t}]=\frac{1}{(s-1)^{2}}

So, the transformation is equal to:

s²·Y(s)+Y(s)=\frac{1}{(s-1)^{2}}

(s²+1)·Y(s)=\frac{1}{(s-1)^{2}}

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

To be able to separate in terms, we use the partial fraction method:

\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

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

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[\frac{s}{2(s^{2} +1)}]-ℒ⁻¹[\frac{1}{2(s-1)}]+ℒ⁻¹[\frac{1}{2(s-1)^{2}}]

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

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3 years ago
A veterinarian knows that a 50-pound dog gets 0.5 milligrams of certain medicine and that the number of milligrams, m, varies di
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The veterinary should give the dog 0.1 m if medicine
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MakcuM [25]

Answer:

C (x+2)(2x+3)

Step-by-step explanation:

You can count in the picture that in the top part there are two x's (2x) and three ones (3). So the top would be (2x+3)

On the left side there is one x (x) and two ones (2). It would be (x+2).

So combine them together, it would be (x+2)(2x+3).

C.

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a course meets 5 times a week. one student misses class 3 days apart. a second student misses class 5 days apart. they were both
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They will both be absent on the 17th day
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