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3 years ago

The expression 12xyz-45 is a a. monomial b. binomial c. trinomial d. constant

Mathematics

1 answer:

ANEK [815]3 years ago

4 0

Answer:

Hey there!

12xyz-45 is a binomial. A binomial contains two terms.

Let me know if this helps :)

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Which graph decreases, crosses the y-axis at (0,-7), and then remains constant?

sergey [27]

Answer:

A.Graph B

Step-by-step explanation:

it decreases to a y of (0,-7) and then stays constant at -7

Hope this was helpful

7 0

3 years ago

Is 5/6 close to 1, 0, or 1/2. Explain

Brums [2.3K]

5/6 is closest to 1 because it it 1/6 away from 1 and is 5/6 away from 0 and is 2/6 away from 1/2 (referring 1/2 as 3/6) Hope this helps!

5 0

3 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 :

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

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

3 years ago

Which statements are true? Check all that apply.

Bumek [7]

Answer:

1,2,4

Step-by-step explanation:

A chord of a circle is a straight line segment whose endpoints both lie on the circle.

KJ is CHORD.

Tangency is point where tangent of circle touch it.

F is point of Tangency for FE.

Secant is the infinite line extension of a chord.

GN is a secant segment. 

4 0

3 years ago

Read 2 more answers

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

$y_2'' + y_2' = -4cos2x-2sin2x$

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

$y_3'' + y_3' = -4sin2x - 2cos2x$

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

6 0

3 years ago