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2 years ago

14

HELP FAST PLEASE The volume of a paper bag with a base of 12" by 5" , and a height of 21" is in cubic inches

1 answer:

Oliga [24]2 years ago

7 0

Answer:1260 in ^3

Step-by-step explanation:

Volume of cubiod= l x b x h

L=12

B= 5

H=21

Volume = 12 x5 x 21

Volume = 1260 in ^ 3

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Andrew is a realtor who makes 11% commissions. if he recently sold a house for $120,000 how much would he make in commissions

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Answer:

the answer would be 133200 dollars

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2 years ago

Taylor baked a cake. Suppose she cut the cake into 8 equal size pieces and 6 people ate all the pieces. Explain how they could h

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3 years ago

Stephen & Richard share a lottery win of £2950 in the ratio 2 : 3. Stephen then shares his part between himself, his wife &a

horsena [70]

Answer:

Stephen's wife got £354 more than his son.

Step-by-step explanation:

Given:

Amount of Lottery = £2950

Now Given:

Stephen & Richard share a lottery amount in the ratio 2 : 3

Let the common factor between them be 'x'.

So we can say that;

$2x+3x=2950\\\\5x = 2950$

Dividing both side by 5 we get;

$\frac{5x}{5}=\frac{2950}{5}\\\\x = 590$

So we can say that;

Stephen share would be = $2x =2\times 590 = \£1180$

Now Given:

Stephen then shares his part between himself, his wife & their son in the ratio 3 : 5 : 2.

Let the common factor between them be 'y'.

So we can say that;

$3y+5y+2y=1180\\\\10y=1180$

Dividing both side by 10 we get;

$\frac{10y}{10}=\frac{1180}{10}\\\\y=118$

So Stephen's wife share = $5y = 5\times 118= \£590$

And Stephen's son share = $2y=2\times118 =\£236$

Now we need to find how much more her wife got then her son.

To find how much more her wife got than her son we will subtract Stephen's son share from Stephen's wife share.

framing in equation form we get;

Amount more her wife got than her son = $590-236 = \£354$

Hence Stephen's wife got £354 more than his son.

3 0

3 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

Simplify (m^3)^6/m^18

Nezavi [6.7K]

Step-by-step explanation:

b = 1........... .. ...... .

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2 years ago