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3 years ago

-4х + 9y = 9 x - Зу = -6

Mathematics

2 answers:

Sati [7]3 years ago

8 0

Answer: x = 9 y = 5

Step-by-step explanation: I like to use elimination for problems like these. Start by multiplying everything in the second equation (x - 3y = -6) by 4. Your equation should then look like this: 4x - 12y = -24

Now add the two equations as if they were two numbers:

-4x + 9y = 9

4x - 12y = -24

Your x's will cancel out and 9y + -12y = -3y and 9 + -24 = -15. This will make your new equation look like:

-3y = -15

Divide both sides by -3 and you get that y = 5

Substitute 5 in for y for any of the original given equations and solve:

-4x + 9y = 9

-4x + 9(5) = 9

-4x + 45 = 9

-4x = -36

x = 9

Flauer [41]3 years ago

5 0

Answer:

x= 9 y = 5, your welcome

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defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

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I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

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$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

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