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3 years ago

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Joanne has a rectangular kitchen that is 10 ft by 15 ft. She decides to increase the length and the width by the same amount (x)

in order to have a new kitchen whose area is double the area of the original kitchen. How much should the length and the width increased by?

1 answer:

Elodia [21]3 years ago

8 0

Answer:

NO

Step-by-step explanation:

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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

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

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

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

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

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2 years ago

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soldi70 [24.7K]

N-10=32
add 10 to both sides
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n=42

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3 years ago

43 grams of copper and 7grams other metals in a birdhouse, 83 grams of copper and 17 grams of other metals. Which one has more c

andrew-mc [135]

Answer: cage I

Step-by-step explanation:

Given

43 gm of copper and 7 gram of other metal

Another cage has 83 gram of copper and 17 gm of other metals

Here, the percentage of copper is

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For the second case

$\Rightarrow \dfrac{83}{83+17}\times 100=83\%$

So, the cage-I has more percentage of copper

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9966 [12]

10000ft<span>² for each floor.

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alexgriva [62]

Answer:

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Step-by-step explanation:

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