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2 years ago

Find a b if a = 2,-8 and b -1, 4

Mathematics

1 answer:

Blababa [14]2 years ago

8 0

Answer:

Step-by-step explanation:

a.b=(2)(-1)+(-8)(4)=-2-32=-34

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Evaluate the expression for the given value of the variable.

Fudgin [204]

Answer:

The answer is three please give brainliest:D

Step-by-step explanation:

4 0

1 year 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 :

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

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

2 years ago

Can anyone help me out, if you answer the question right i will give you a brainliest.

melisa1 [442]

Answer:

It's different because the experiment is more accurate as it progresses.

Step-by-step explanation:

You'll notice that the higher the numbers get in the experiment the closer it gets to your solution. The theoretical probability of flipping a coin is about 50% heads and 50% tails, but it doesn't always seem like that in an experiment. The experimental probability from your experimentation so far would be 62% of heads and 38% of tails.

4 0

2 years ago

((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2

Vitek1552 [10]

Answer:

256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

(2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

(a^b)^c = a^(b·c)

(a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

(2^2)(2^((-3)(-2)) = (2^2)(2^6)

= 2^(2+6)

= 2^8 = 256

3 0

2 years ago

Find the equation of the line that passes through points A and B. y X A (4, 7) B (2, 3)

g100num [7]

Answer:

y = 2x - 1.

Step-by-step explanation:

The slope = (7-3)/(4-2)

= 4/2

= 2.

y - y1 = m(x - x1)

Here m = 2 , x1 = 2 and y1 = 3. So we have:

y - 3 = 2(x - 2)

y = 2x - 4 + 3

y = 2x - 1.

We have used the point (2, 3) to find the equation but we could have used (4, 7). We would have got the same answer.

3 0

2 years ago

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