Factor 16x2 + 16x + 4 completely.

16x^2 + 16x + 4 = 4(1 + 4x + 4x^2) = 4(2x + 1)^2

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Asked 1/26/2011 3:28:49 PM

Updated 8/9/2014 9:22:43 AM

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16x^2 + 16x + 4 = 4(1 + 4x + 4x^2) = 4(2x + 1)^2

Added 8/9/2014 9:22:43 AM

This answer has been confirmed as correct and helpful.

Confirmed by yumdrea [8/9/2014 9:28:03 AM]

A circle has a diameter with endpoints of (-3, 8) and (7, 4). What is the center of the circle? **Weegy:** (2, 6). **User:** Find the length of the hypotenuse of a right triangle whose legs are 5 and abs(2). (More)

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Expert Answered

Updated 123 days ago|8/2/2023 12:46:47 AM

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To find the length of the hypotenuse in a right triangle, you can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs.

Given that the lengths of the legs are 5 and abs(2) (which is also 2), we can substitute these values into the Pythagorean theorem:

Hypotenuse^2 = Leg1^2 + Leg2^2

Hypotenuse^2 = 5^2 + 2^2

Hypotenuse^2 = 25 + 4

Hypotenuse^2 = 29

To find the length of the hypotenuse, we need to take the square root of both sides:

Hypotenuse = sqrt(29)

Using a calculator, we find that the square root of 29 is approximately 5.385.

Therefore, the length of the hypotenuse in the right triangle with legs 5 and abs(2) is approximately 5.385 units.

Given that the lengths of the legs are 5 and abs(2) (which is also 2), we can substitute these values into the Pythagorean theorem:

Hypotenuse^2 = Leg1^2 + Leg2^2

Hypotenuse^2 = 5^2 + 2^2

Hypotenuse^2 = 25 + 4

Hypotenuse^2 = 29

To find the length of the hypotenuse, we need to take the square root of both sides:

Hypotenuse = sqrt(29)

Using a calculator, we find that the square root of 29 is approximately 5.385.

Therefore, the length of the hypotenuse in the right triangle with legs 5 and abs(2) is approximately 5.385 units.

Added 150 days ago|7/6/2023 12:03:44 AM

To find the center of the circle, we can use the midpoint formula, which states that the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is ((x1 + x2) / 2, (y1 + y2) / 2).

Let's apply the formula to the given endpoints (-3, 8) and (7, 4):

Center = ((-3 + 7) / 2, (8 + 4) / 2)

Center = (4 / 2, 12 / 2)

Center = (2, 6)

So, the center of the circle is at the point (2, 6).

Let's apply the formula to the given endpoints (-3, 8) and (7, 4):

Center = ((-3 + 7) / 2, (8 + 4) / 2)

Center = (4 / 2, 12 / 2)

Center = (2, 6)

So, the center of the circle is at the point (2, 6).

Added 123 days ago|8/2/2023 12:46:47 AM

This answer has been confirmed as correct and helpful.

true or false is this a quadratic equation. y(y + 4) - y^2 = 6 **Weegy:** -11y + 3 (y - 10) = -11y + 3y - 30 = -9y - 30
craig1990|Points 205|
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Updated 130 days ago|7/26/2023 9:50:33 AM

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y(y + 4) - y^2 = 6 is a quadratic equation. TRUE

Added 130 days ago|7/26/2023 9:50:33 AM

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What is range? **Weegy:** The range of a set of numbers is the highest number minus the lowest number.
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Updated 3/29/2016 1:40:45 PM

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Find the midpoint of the line segment whose endpoints are (-2, 5) and (6, -9). **Weegy:** Whole Numbers are simply the numbers 0, 2, 4, 6. Integers are like whole numbers, but they also include negative numbers -8, -6, -4, -2, 0, 2, 4, 6 [ ] (More)

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Updated 7/27/2014 1:01:12 PM

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The midpoint of the line segment whose endpoints are (-2, 5) and (6, -9) is (2, -2). M = [(-2+6/2), (5-9/2)]; M = (4/2, -4/2); M = (2, -2)

Added 7/27/2014 1:01:11 PM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [7/27/2014 1:03:28 PM]

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