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3 years ago

( 1,-2), gradient = -3

Mathematics

2 answers:

igor_vitrenko [27]3 years ago

6 0

Answer:

if you are required to find the equation of a straight line use the formula y-y1= m (x-x1)

y+2=-3(x-1)

y+2=-3x+3

y= -3x+3-2

y -3x+1

hope this helps

Nesterboy [21]3 years ago

5 0

Answer:

y = -3x + 1

Step-by-step explanation:

(y -(-2)) = -3(x-1)

y+ 2 = -3x+ 3

y = -3x + 1

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Nutka1998 [239]

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3 years ago

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Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

solniwko [45]

For this case we have to:

Let $u = x + 1$

So:

$u ^ 2-4u + 2 = 0$

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Where:

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Substituting:

$u = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (1) (2)}} {2 (1)}\\u = \frac {4 \pm \sqrt {16-8}} {2}\\u = \frac {4 \pm \sqrt {8}} {2}\\u = \frac {4 \pm \sqrt {2 ^ 2 * 2}} {2}\\u = \frac {4 \pm2 \sqrt {2}} {2}$

The solutions are:

$u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}$

Returning the change:

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Answer:

$x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}$

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3 years ago

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Pepsi [2]

Hello!

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You can plug in the values you know

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4 years ago

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sdas [7]

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Answer:

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3 years ago

( 1,-2), gradient = -3​

( 1,-2), gradient = -3