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madreJ [45]
3 years ago
15

justin was trying to figure out how to write the expression 7 less than 'the double of twenty five. help justin write the expres

sion correctly, and then solve.
Mathematics
2 answers:
Amanda [17]3 years ago
7 0
The double of 25 is (25+25), or (25x25). 7 less would be (25+25) -7. Figure out what 25 + 25 is, then subtract 7 from your answer.
vova2212 [387]3 years ago
7 0
25 x 2 -7 = 43.

Hi! The “7 less than” would immediately go to the back because it is subtracting from something. The double means to multiply the 25 times two. You wouldn’t need parenthesis because of the rules of PEMDAS. 50-7 = 43.

Hope this helps!
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Answer:

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(-13) -Y

putting the value of y = 7 in the equation

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A marathon race participant walks at a rate of 2.5 miles per hour and runs at a rate of 5 miles per hour. The combined distance
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1/4x + 3/4 = 1/2 ??????????
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Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2
kifflom [539]

Answer:

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)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)=e^{rt}, where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

y'=re^{rt}

y"=r^{2}e^{rt}

r^{2}e^{rt}+e^{rt}=0

(r^{2}+1)e^{rt}=0

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

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}

Using Euler's formula:

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

<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)]=\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)

3 0
3 years ago
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