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1 year ago

Consider the following system of two linear equations:

Mathematics

1 answer:

harkovskaia [24]1 year ago

4 0

Hello,

Answer:

The last graph (2;3)

Step-by-step explanation:

2x + 3y = 12

2x – 3y = 0

2x + 3y = 12

2x = 3y

3y + 3y = 12 ⇔ 6y = 12 ⇔ y = 12/6 = 2

2x + 3 × 2 = 12 ⇔ 2x + 6 = 12 ⇔ 2x = 6 ⇔ x = 6/2 = 3

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An athlete trains for 95 min each day for as many days as possible.Write an equation that relates the number of days d that the

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855=95d

855 minutes is equal to 95 minutes times the number of days run

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3 years ago

Productive use of machinery on a farm made it possible to decrease the number of tractors by 12. How many tractors does the farm

frez [133]

1. the answer is 24. think of x as the original amount, and y as the new amount. y times 1.5 is x, and y+12 is x. reverse that to figure out y, which is what we need, and you have x/1.5 = y as well as x-12 = y. Use the equal values method and make an equation x/1.5=x-12. solve this equation to get x, which is 36. to figure out the new amount, y, you need to subtract 12, which would help you get 24 as your final answer.

2. once again, create an equation. let's call team 1 x and team 2 y. team one has 1/4th as many as team 2, so that would be x=1/4y. An easier way to write that is 4x=y. after 6 people quit team two, that would be y-6. after the transfer, that would be y-6-12, and x+12 for the teams. they are equal after these, so y-6-12=x+12. solve this equation to get y-18= x+12. if you recall earlier, y was 4 times x, so substitute that into y to get 4x-18=x+12. Solve the equation to get 10 people on team one originally. your final answer is 10 people.

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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

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

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3 years ago

Find from u. -12u = -24

Bingel [31]

Well, to find u, we have to remove all that is attached to it so the equation can just be u=...
To find u, you have to remove what is attached to it, and that is -12. Then you have to look at the relationship between the -12 and u. The relationship is multiplication, and the opposite of multiplication is division, so all you have to do is divide both sides by -12. So;
-12u/-12=-24/-12
The -12 cancels the -12, leaving u and the - in 12 cancels the - in 24. Leaving 24/12. And that is 2. Written as;
u=2
Hope i helped. If you have any more problems, let me know.

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3 years ago

Read 2 more answers

Given the equations 9x+3/4y=6 and 2x+1/2y=9, by what factor the equation to eliminate y and solve the system through linea would

lions [1.4K]

C -3/2

Just multiply the second equation by -3/2
this will make the y term = -3/4y

so adding will eliminate the y terms.

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3 years ago

Read 2 more answers