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gtnhenbr [62]
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

Mathematics
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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vova2212 [387]

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
Karolina [17]
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5 0
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........... .. ...... .

7 0
2 years ago
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