39 points no question

Mathematics

2 answers:

aliya0001 [1]3 years ago

7 0

Answer:

thanks

Step-by-step explanation:

you are great hehehejeieii3i3j3

DaniilM [7]3 years ago

6 0

Answer:

Ohhh ok No problem i will accept it

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Two lines are perpendicular. The equation of the first line is y = 4x + 1. Which of the following cannot be the equation of the

zalisa [80]

The answer is letter B

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3 years ago

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zysi [14]

Answer:

Step-by-step explanation:

$\frac{a^2+6a+9}{a^2-9} *\frac{3a-9}{a+3} \\=\frac{a^2+2*a*3+3^2}{a^2-3^2} *\frac{3(a-3)}{a+3} \\=\frac{(a+3)^2}{(a+3)(a-3)} *\frac{3(a-3)}{a+3} \\=\frac{3(a+3)^2}{(a+3)^2} \\=3$

$\frac{3x^2-6x}{3x+1} *\frac{x+3x^2}{x^2-4x+4} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{x^2-2x-2x+4} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{x(x-2)-2(x-2)} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{(x-2)(x-2) } \\=\frac{3x^2}{x-2)}$

$\frac{2x^2-10x+12}{x^2-4} *\frac{2+x}{3-x} \\=\frac{2[x^2-5x+6]}{x^2-2^2} *\frac{2+x}{-(-3+x)} \\=\frac{2[x^2-2x-3x+6]}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2[x(x-2)-3(x-2)]}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2(x-2)(x-3)}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2}{-1} \\=-2$

$\frac{6x^2-11x-10}{6x^2-5x-6} *\frac{6-4x}{25-20x+4x^2} \\=\frac{6x^2-15x+4x-10}{6x^2-9x+4x-6} *\frac{-4x+6}{4x^2-10x-10x+25} \\=\frac{3x(2x-5)+2(2x-5)}{3x(2x-3)+2(2x-3) } *\frac{-2(2x-3)}{2x(2x-5)-5(2x-5)} \\=\frac{(2x-5)(3x+2)}{(2x-3)(3x+2)} *\frac{-2(2x-3)}{(2x-5)(2x-5)} \\=\frac{-2}{2x-5}$