Im so confused (please hurry and answer this is due soon

Mathematics

1 answer:

frozen [14]3 years ago

3 0

Answer:

its 12 :)

hope this helps you

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If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

So, no option is correct. And the answer is

$T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

Learn more about linear transformations here-

brainly.com/question/13005179

#SPJ10

5 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$