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nignag [31]
3 years ago
15

How to solve number 1?

Mathematics
1 answer:
Archy [21]3 years ago
5 0
E(x)=5(x-1)(x+2)-(x+2)^2\\\\1)\ 5(x^2+2x-x-2)-(x^2+2x\cdot2+2^2)\\=5(x^2+x-2)-(x^2+4x+4)\\=5x^2+5x-10-x^2-4x-4\\=4x^2+x-14\neq4x^2+x+1\\\\2)\ E(x)=1\ for\ E(x)=4x^2+x+1\\\\4x^2+x+1=1\ \ \ \ |subtract\ 1\ from\ both\ sides\\4x^2+x=0\\x(4x+1)=0\iff x=0\ \vee\ 4x+1=0\\x=0\ \vee\ 4x=-1\\\boxed{x=0\ \vee\ x=-0.25}

3)\ H(x)=9x^2-(2x+1)^2\\\\a.\ H(x)=(3x)^2-(2x+1)^2=[3x-(2x+1)](3x+2x+1)\\=(3x-2x-1)(5x+1)=(5x+1)(x-1)\\\\b.\ H(x)=0\\\\(5x+1)(x-1)=0\iff5x+1=0\ \vee\ x-1=0\\5x=-1\ \vee\ x=1\\\boxed{x=-0.2\ \vee\ x=1}
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11 feet = 132 inches
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6 0
4 years ago
I dont mind giving out 40 points to who ever answers this correctly
jenyasd209 [6]

Answer:

27/ (4  x^6 y^8)

Step-by-step explanation:

4(3x^2y^4)^3/(2x^3y^5)^4

expand

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power of the power rule , exponents to the power are multiplied

4 (3^3  x^(2*4) y^(4*3) / (2^4 x^(3*4) y^(5*4))

4 (27  x^(8) y^(12) / (16 x^(12) y^(20))

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when dividing exponents , we subtract the powers

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27/4  *  x^(-6)(y)^ (-8)

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5 0
4 years ago
Can anyone help me with this?
Artemon [7]

7) 5+4 = 9

4+5=9

8) 6+5 = 11

5+6 = 11

9) 6+7 = 13

7+6 = 13

10) 7+8 = 15

8+7 = 15

11) 8+9 = 17

9+8 = 17

12) 5+6 = 11

6+5 = 11

13) 7+6 = 13

6+7 = 13

14) 9+8 = 17

8+9 = 17

3 0
3 years ago
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Mademuasel [1]

Answer:

3

Step-by-step explanation:

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