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GREYUIT [131]
3 years ago
7

Plese someone answer me b asap

Mathematics
2 answers:
Oduvanchick [21]3 years ago
8 0

Answer:

\huge\boxed{\sf 9y^2 -22y-24}

Step-by-step explanation:

\sf (3y-1)^2 -(2y+4)^2+(2y-3)(2y+3)\\\\Using \ formula:\\\\1) (a-b)^2 = a^2-2ab+b^2\\\\2) (a+b)^2= a^2+2ab+b^2\\\\3) (a+b)(a-b) = a^2-b^2\\\\Applying \ Formulas\\\\(3y)^2-2(3y)(1)+(1)^2-[(2y)^2+2(2y)(4)+(4)^2]+(2y)^2-(3)^2\\\\9y^2-6y+1-[4y^2+16y+16]+4y^2-9\\\\9y^2-6y+1 -4y^2-16y-16+4y^2-9\\\\Combining \ like \ terms\\\\9y^2-4y^2+4y^2-6y-16y+1-16-9\\\\9y^2 -22y-24\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
Ipatiy [6.2K]3 years ago
3 0

Answer:

a) \frac{1}{3} <em>k</em>² - 9

b) -7<em>y</em>² - 3

Step-by-step explanation:

a) (\frac{2}{3} <em>k</em> - 6)² - (\frac{1}{3} <em>k </em>+ 3)²

  = \frac{4}{9} <em>k</em>² - 36 - \frac{1}{9} <em>k</em>² + 9

  = \frac{4}{9} <em>k</em>² - \frac{1}{9} <em>k</em>² + 9 -36

  = \frac{3}{9} <em>k</em>² - 27

  = \frac{1}{3} <em>k</em>² - 9

____________________________________

b) (3<em>y -</em> 1)² - (2<em>y</em> + 4)² + (2<em>y - </em>3) (2<em>y</em> + 3)

  = 9<em>y</em>² - 1 - 4<em>y</em>² + 16 + (2<em>y - </em>3) (2<em>y</em> + 3)

  = 9<em>y</em>² - 4<em>y</em>² - 1 + 16 + (<em>- </em>3) (4<em>y</em>² + 6<em>y</em>)

  = 5<em>y</em>² + 15 - 12<em>y</em>² - 18

  = 5<em>y</em>² - 12<em>y</em>² + 15 - 18

  = -7<em>y</em>² - 3

Hope this helps!

(NOTE: Do you mind marking me as Brainliest?)

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