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Burka [1]
3 years ago
10

A dilation is given by the transformation (x, y) - (.75x, 75y). What type of dilation is this

Mathematics
1 answer:
uranmaximum [27]3 years ago
3 0
Most likely B because x>1 = enlargement vs x<1 = reduction, & the x & y are doing two completely different things
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Answer:

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Step-by-step explanation:

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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.

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3 years ago
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Cloud [144]

Answer:

Lucy cuts 4 squares with side length x in. from the corners of a 6 in. by 9 in. cardboard rectangle. She folds the remaining cardboard to make a tray that is x in. high. Write and simplify a polynomial function for the volume V

Step-by-step explanation:

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expanded form is adding each place value together with zeros replacing all other spots

32.28

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