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german
3 years ago
9

A homeowner has 20 feet of fencing material to enclose a rectangular area for his pets. the rectangular area is adjacent to a ho

use. the side that is adjacent to the house does not need to have railing installed. what dimensions should be used in order to maximize the play area?
Mathematics
1 answer:
VikaD [51]3 years ago
7 0
5X10=50 sq ft area
Around the 3 sides= 5+10+5=20
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5. Enter the answer as a decimal
Hitman42 [59]

Answer:

6.5%

Step-by-step explanation:

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Triangle ABC Is similar to triangle DEF solve for r
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since they're similair then similair sides will be proportional

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same thing if you try with the hypotenuse AB similair to DE

\frac{13.5}{18}1813.5 = \frac{3}{4}43

So, if you multiply  r (DF) by 3/4  you will get the similair side AC

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<h3>r=16.</h3>

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4 0
2 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

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Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

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Integrating factor

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Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
2 years ago
Work out these approximate conversions.
Anvisha [2.4K]

Answer:

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Step-by-step explanation:

5 0
2 years ago
Please help me I am struggling
Oxana [17]

Answer:

it's 48

Step-by-step explanation:

Because 8x3=24 and you multiply 6×8=48

3 0
2 years ago
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