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2 years ago

5

Evaluate.

1 answer:

lys-0071 [83]2 years ago

8 0

Answer:

361/900

Step-by-step explanation:

$\left(-\dfrac{1}{6}+0.6\left(-\dfrac{1}{3}\right)+1\right)^2=\left(-\dfrac{1}{6}-0.2+1\right)^2\\\\=\left(1-\left(\dfrac{1}{6}+\dfrac{1}{5}\right)\right)^2=\left(1-\dfrac{5+6}{6\cdot5}\right)^2=\left(\dfrac{19}{30}\right)^2=\boxed{\dfrac{361}{900}}$

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3 years ago

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Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

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So we can say that

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Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

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it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

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3 years ago