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ad-work [718]
3 years ago
13

How to solve to system y=x^2-2 y=2x+1

Mathematics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

the solutions are (3, 7) and (-1, -1)

Step-by-step explanation:

Insert the " = " symbol between the two equations, obtaining:

y = x^2 - 2 = y = 2x + 1

Then x^2 - 2 = 2x + 1, or

x^2 - 2x - 3 = 0, and this can be factored into (x - 3)(x + 1) = 0.

Thus, the x values that satisfy this system are {-1, 3}.

Use y = 2x + 1 to find the corresponding y values:

y = 2(-1) + 1 = -1

and

y = 2(3) + 1 = 7

Then the solutions are (3, 7) and (-1, -1)

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Archy [21]

The value of integration of y=16-x^{2} from x=-1 to x=1 is 94/3.

Given the equation y=16-x^{2} and the limit of the integral be x=-1,x=1.

We are required to find the value of integration of y=16-x^{2} from x=-1 to x=1.

Equation is relationship between two or more variables that are expressed in equal to form.Equation of two variables look like ax+by=c.It may be linear equation, quadratic equation, or many more depending on the power of variable.

Integration is basically opposite of differentiation.

y=16-x^{2}

Find the integration of 16-x^{2}.

=16x-x^{3}/3

Now find the value of integration from x=-1 to x=1.

=16(1)-(1)^{3}/3-16(-1)-(-1)^{3}/3

=16(1)-1/3+16-1/3

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Hence the value of integration of y=16-x^{2} from x=-1 to x=1 is 94/3.

Learn more about integration at brainly.com/question/27419605

#SPJ4

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