Draw the image of \triangle ABC△ABCtriangle, A, B, C under a dilation whose center is PPP and scale factor is \dfrac{1}{3} 3 1
1 answer:
Dilating the triangle ABC would create another triangle whose size is different from the triangle ABC
<h3>How to dilate the triangle?</h3>From the complete question, we have the following coordinates:
- Point A = (-4,5)
- Point B = (7,-7)
- Point C = (-2,2)
- Point P = (1,-4)
- Scale factor, k = 1/3
The dilation rule across the center P is:
(x, y) --> (1/3x – 1, 1/3y + 4)
So, the coordinates of the image of the triangle are:
A' = (-1/3 * 4 – 1, 1/3 * 5 + 4)
A' = (-7/3, 17/3)
B' = (1/3 * 7 – 1, -1/3 * 7 + 4)
B' = (4/3, 5/3)
C' = (-1/3 * 2 – 1, 1/3 * 2 + 4)
C' = (-5/3, 14/3)
From the above computation, the coordinates of the new triangle is:
A' = (-7/3, 17/3)
B' = (4/3, 5/3)
C' = (-5/3, 14/3)
See attachment for the image of triangle ABC
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Answer:
The solution of the differential equation is
Step-by-step explanation:
The differential equation is given by: y" + y = Sin(2t)
<u>i) Using characteristic equation:</u>
The characteristic equation method assumes that y(t)=, where "r" is a constant.
We find the solution of the homogeneus differential equation:
y" + y = 0
As could never be zero, the term (r²+1) must be zero:
(r²+1)=0
r=±i
The solution of the homogeneus differential equation is:
Using Euler's formula:
The particular solution of the differential equation is given by:
So we use these derivatives in the differential equation:
As there is not a term for Cos(2t), B is equal to 0.
So the value A=-1/3
The solution is the sum of the particular function and the homogeneous function:
Using the initial conditions we can check that C1=5/3 and C2=2
<u>ii) Using Laplace Transform:</u>
To solve the differential equation we use the Laplace transformation in both members:
ℒ[y" + y]=ℒ[Sin(2t)]
ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]
By using the Table of Laplace Transform we get:
ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1
ℒ[y]=Y(s)
ℒ[Sin(2t)]=
We replace the previous data in the equation:
s²·Y(s)
-2s-1+Y(s)
=
(s²+1)·Y(s)-2s-1=
(s²+1)·Y(s)=
Y(s)=
Y(s)=
Using partial franction method:
2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)
2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)
We solve the equation system:
A+C=2
B+D=1
A+4C=8
B+4D=6
The solutions are:
A=0 ; B= -2/3 ; C=2 ; D=5/3
So,
Y(s)=
Y(s)=
By using the inverse of the Laplace transform:
ℒ⁻¹[Y(s)]=ℒ⁻¹[]-ℒ⁻¹[
]+ℒ⁻¹[
]