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oksano4ka [1.4K]
3 years ago
5

HELP PLEASE! Write an equation with variables on both sides that has a single solution "-1." Explain how to find that solution

Mathematics
2 answers:
Inessa05 [86]3 years ago
8 0

ANSWER : HEY MATE

Art [367]3 years ago
3 0
Can be like that? The answer will be always -1 because of the negative number in the secon term

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Saturday = 46+19 min = 65 minutes
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3 years ago
Need help factoring this binomial: x^4-1?
natima [27]
Use the difference of squares factorization - that for any numbers a and b, (a-b)(a+b)=a^2-b^2.

We have:

(x^2+1)(x^2-1)=x^4-1

In addition:

(x-1)(x+1)=x^2-1, so we have:

(x^2+1)(x+1)(x-1)

As our complete factorization.
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Do imaginary numbers follow the rules of algebra?
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Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?
Lubov Fominskaja [6]

Answer:

  yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

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For the first parabola, the point of intersection is ...

  y^2 = 4(-y-1)

  y^2 +4y +4 = 0

  (y+2)^2 = 0

  y = -2 . . . . . . . . one solution only

  x = -(-2)-1 = 1

The point of intersection is (1, -2).

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For the second parabola, the equation is the same, but with x and y interchanged:

  x^2 = 4(-x-1)

  (x +2)^2 = 0

  x = -2, y = 1 . . . . . one point of intersection only

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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.

7 0
3 years ago
Find the slope of the line that passes through (8, 6) and (2, 7).
Amiraneli [1.4K]
Vas happenin!!

Slope form - y2-y1/x2-x1

7-6/ 2-8

1/-6

Your slope is -6x


Hope this helps

-Zayn Malik
6 0
3 years ago
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