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3 years ago

Simplify (4 1/3)3 x (61/6)6 x (3 1/2)4

Mathematics

1 answer:

VMariaS [17]3 years ago

3 0

Answer:

11102 x^2

Step-by-step explanation:

Simplify the following:

((4 + 1/3)×3×61×6 (3 + 1/2)×4 x x)/6

61/6×6 = (61×6)/6:

(4 + 1/3)×3×(61×6)/6 (3 + 1/2)×4 x x

((3 + 1/2)×4×61×6 (4 + 1/3)×3 x x)/6 = 6/6×(3 + 1/2)×4×61 (4 + 1/3)×3 x x = (3 + 1/2)×4×61 (4 + 1/3)×3 x x:

(3 + 1/2)×4×61 (4 + 1/3)×3 x x

Put 4 + 1/3 over the common denominator 3. 4 + 1/3 = (3×4)/3 + 1/3:

(3 + 1/2)×4×61 (3×4)/3 + 1/3 3 x x

3×4 = 12:

(3 + 1/2)×4×61 (12/3 + 1/3)×3 x x

12/3 + 1/3 = (12 + 1)/3:

(3 + 1/2)×4×61 (12 + 1)/3×3 x x

12 + 1 = 13:

((3 + 1/2)×4×61×13×3 x x)/3

Put 3 + 1/2 over the common denominator 2. 3 + 1/2 = (2×3)/2 + 1/2:

(4×61×13×3 x x)/3 (2×3)/2 + 1/2

2×3 = 6:

((6/2 + 1/2)×4×61×13×3 x x)/3

6/2 + 1/2 = (6 + 1)/2:

(4×61×13×3 x x)/3 (6 + 1)/2

6 + 1 = 7:

(7×4×61×13×3 x x)/(2×3)

(7×4×61×13×3 x x)/(2×3) = 3/3×(7×4×61×13 x x)/2 = (7×4×61×13 x x)/2:

(7×4×61×13 x x)/2

4/2 = (2×2)/2 = 2:

7×2×61×13 x x

7×2×61×13 x x = 7×2×61×13 x^2:

7×2×61×13 x^2

7×2 = 14:

14×61×13 x^2

14×61 = 854:

854×13 x^2

854×13 = 11102:

Answer: 11102 x^2

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The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

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As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

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$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

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:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

ii) Using Laplace Transform:

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)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

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)= $\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}$

Y(s)= $-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

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3 years ago