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schepotkina [342]
3 years ago
8

If x^2 - 16 > 0, which of the following must be true?

Mathematics
1 answer:
zimovet [89]3 years ago
5 0
The left hand side of the equation is a difference of two squares and may be factored out as follows,
                              (x - 4)(x + 4) > 0
They may be individually taken as,
                                     x - 4 > 0    ; x > 4
                                     x + 4 > 0   ; x < -4
Thus, the answer to this item is letter A. 
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The equation of the axis of symmetry is x = 3. True or false
Vsevolod [243]

Answer:

true

Step-by-step explanation:

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Use the commutative property to rewrite the expression x + 16 *<br><br><br> please help me
valentinak56 [21]

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16+x

3 0
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arman needs 4 1/2 inches of cable to make A,3 3/4 inches to make B, and 5 1/8 inches to make C. He used a 12 inch piece of cable
Kaylis [27]
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He used 12 inches of cable...
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6 0
4 years ago
Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.
solniwko [45]

For this case we have to:

Letu = x + 1

So:

u ^ 2-4u + 2 = 0

We have the solution will be given by:

u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Where:

a = 1\\b = -4\\c = 2

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:

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

Answer:

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

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3 years ago
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Aleks04 [339]
I will take110+110 which will give me220
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