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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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The correct answer is:

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\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}

The lengths of the sides of KELY are:

\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}

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Answer:

a) Q(-2,1) is false

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Step-by-step explanation:

Given data is Q(x,y) is predicate that x then x^{2}. where x,y are rational numbers.

a)

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4 this is wrong. since 4>1

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b)

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25 this is not true. since 25>4.

This is similar to the truth value of part (a).

Since in both x satisfied and x^{2} >y^{2} for both the points.

c)

if Q(x,y)=Q(3,8) that is x=3 and y=8

Here 3 this satisfies the condition x.

Then 3^{2}

9 This also satisfies the condition x^{2}.

Hence Q(3,8) exists and it is true.

d)

Assume Q(x,y)=Q(9,10)

Here 9 satisfies the condition x

Then 9^{2}

81 satisfies the condition x^{2}.

Thus, Q(9,10) point exists and it is true. This satisfies the same values as in part (c)

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