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(6a + b)2

Weegy: 12z + 2b (More)

Question

Updated 94 days ago|6/20/2014 8:39:46 AM

1 Answer/Comment

(6a + b)^2

= (6a + b)(6a + b)

= 36a^2 + 6ab + 6ab + b^2

= 36a^2 + 12ab + b^2

= (6a + b)(6a + b)

= 36a^2 + 6ab + 6ab + b^2

= 36a^2 + 12ab + b^2

Added 94 days ago|6/20/2014 8:39:46 AM

Find the quotient. y9 ÷ y3

Weegy: = y3 (More)

Question

Updated 69 days ago|7/15/2014 2:56:28 AM

1 Answer/Comment

Evaluate y^5 for y = 1.

Weegy: HY HOW MAY I HELP YOU User: Evaluate y5 for y = 1? User: can you evaluate y^5 for y= 1? (More)

Question

Updated 87 days ago|6/27/2014 4:12:52 PM

1 Answer/Comment

If the equation of a circle is (x + 4)2 + (y - 6)2 = 25, its center point is

Weegy: False.
y(y 4) - y = 6 is written as:
y^5 -y - 6 = 0.
A quadratic equation is a second-order polynomial equation in a single variable y and this is not a second order polynomial equation.
(More)

Question

Updated 81 days ago|7/3/2014 7:48:18 PM

1 Answer/Comment

If the equation of a circle is (x + 4)^2 + (y - 6)^2 = 25, its center point is (-4, 6).

Added 81 days ago|7/3/2014 7:48:18 PM

What is the solution set of (x - 2)(x - 3) = 2?

Question

Updated 119 days ago|5/26/2014 8:16:37 PM

1 Answer/Comment

(x - 2)(x - 3) = 2

x^2 - 3x - 2x + 6 = 2

x^2 - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1 or x = 4

x^2 - 3x - 2x + 6 = 2

x^2 - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1 or x = 4

Added 119 days ago|5/26/2014 8:16:37 PM

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