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2 years ago

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Which represents the solution(s) of the graphed system of equations, y = x2 + x – 2 and y = 2x – 2? (–2, 0) and (0, 1) (0, –2) a

nd (1, 0) (–2, 0) and (1, 0) (0, –2) and (0, 1) Mark this and return

2 answers:

MA_775_DIABLO [31]2 years ago

5 0

Answer:

<h2>(0, -2) (1, 0)</h2>

Step-by-step explanation:

I got it right on the test... Have a great day :)

Luda [366]2 years ago

3 0

ANSWER

$(0,-2), (1,0)$

EXPLANATION

The first equation is

$y = {x}^{2} + x - 2$

The second equation is

$y = 2x - 2$

We equate both equations to get,

${x}^{2} + x - 2 = 2x - 2$

${x}^{2} + x - 2x - 2 + 2 = 0$

Simplify

${x}^{2} - x = 0$

Factor

$x(x - 1) = 0$

Either

$x = 0$

Or

$x - 1 = 0$

$x = 1$

Put x=0 or x=1 into the second equation to get,

$y = 2(0) - 2 = - 2$

Or

$y = 2(1) - 2 = 0$

Therefore the solutions are;

$(0,-2), (1,0)$

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b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

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