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pashok25 [27]
3 years ago
9

The width of a rectangular garden is 9 feet, and its length is 12 feet. Three of these expressions equal the perimeter of the ga

rden, in feet. Which expression does NOT?
9+12
9+12+9+12
2*9+2*12
2(9+12)
Mathematics
2 answers:
soldi70 [24.7K]3 years ago
6 0

Answer:

9+2 is not the expresssion

Step-by-step explanation:

marissa [1.9K]3 years ago
6 0

Answer:

The expression which is does not correct is A. 9 + 12

Step-by-step explanation:

Length of the rectangular garden = 12 feet

Width of the rectangular garden = 9 feet

We need to find the perimeter of the garden.

The perimeter of the garden is given by the formula = 2 × (Length + Width)

⇒ Perimeter = 2 × (12 + 9)

                     = 2 × 12 + 2 × 9

                     = 12 + 12 + 9 + 9

So, All the options B. , C, and D. are correct to find the perimeter of the rectangular field.

Therefore, The expression which is does not correct is A. 9 + 12

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From the figure attached,

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2 years ago
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0
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