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3 years ago

6

3. The positive root of 154 is located between which two consecutive integer values?

2 answers:

telo118 [61]3 years ago

6 0

Answer:

Step-by-step explanation:

12²=144

√144=12

13²=169

√169=13

√154 lies between 12 and 13.

lesantik [10]3 years ago

4 0

Answer:

between 12 and 13

Step-by-step explanation:

answer is 12.4

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2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

3 years ago

Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui

Vesnalui [34]

<h3>The dimensions of the given rectangular box are:</h3><h3>L = 15.874 cm , B = 15.874 cm , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box

= Area of the base + 4 x ( Area of the adjacent faces)

= S x S + 4 ( S x H) = S² + 4 SH ..... (1)

Also, Area of the box = S x S x H = S²H

⇒ S²H = 2000

$\implies H = \frac{2000}{S^2}$

Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S = 15.874 cm in the value of H , we get:

$\implies H = \frac{2000}{S^2} = \frac{2000}{(15.874)^2} = 7.8937 cm$

Hence, the dimensions of the given rectangular box are:

L = 15.874 cm

B = 15.874 cm

H = 7.8937 cm

3 0

3 years ago