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Brut [27]
3 years ago
5

How can I make this a perfect square w^2-4w+

Mathematics
1 answer:
Neporo4naja [7]3 years ago
8 0
Ax^2+bx+c
if a=1, then
take 1/2 of b and square it

w^2-4w+c
b=-4
-4/2=-2, (-2)^2=4
add that
w^2-4w+4
(w-2)(w-2)
(w-2)^2

to make it into perfect squres, add 4
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Compare and contrast the solution of |2x + 8| > 6 and the solution of |2x + 8| < 6.
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Answer:

B.

The solution of |2x + 8| > 6 includes all values that are less than –7 or greater than –1.

The solution of |2x + 8| < 6 includes all values that are greater than –7 and less than –1.

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Step-by-step explanation:

You can find the solution by "unfolding" the absolute value, then dividing by 2 and subtracting 4:

-6 > 2x +8 > 6 . . . . . read this as -6 is less than 2x+8 or 2x+8 is greater than 6

-3 > x +4 > 3 . . . . . . .divide by 2

-7 > x > -1 . . . . . . . . . solution to the first inequality: x is less than -7 or greater than -1.

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The solution to the other inequality is identical, except the direction of the comparison is reversed. It is read differently, because the segments overlap, rather than being disjoint.

-7 < x < -1 . . . . . . . . solution to the second inequality: x is greater than -7 and less than -1.

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These descriptions match choice B.

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