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3 years ago

Which choice answer is it?

Mathematics

1 answer:

Debora [2.8K]3 years ago

8 0

Your answer would be option (1) -3x² - 18x + 19.

To find this we need to substitute in the G and F expressions into G - 3F and see what the outcome is, as this will give us the answer.

First I’m going to multiply the F expression, 2x² + 6x - 5, by 3:

3 × (2x² + 6x - 5) = 3(2x²) + 3(6x) - 3(5) = 6x² + 18x - 15.

Now we need to subtract this expression from G, which is 3x² + 4, which I will do by term:

3x² - 6x² = -3x²

0x - 18x = -18x

4 - -15 = 19

So therefore overall your answer is -3x² - 18x + 19.

I hope this helps! Let me know if you have any questions :)

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Solve irrational equation pls

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$\hbox{Domain:}\\
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x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
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$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
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(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
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(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
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(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
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x-1=0 \vee 3x^2-28=0\\
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x=1 \vee x^2=\dfrac{28}{3}\\
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There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
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\boxed{\boxed{x=1}}$

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3 years ago

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lys-0071 [83]

$\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$

-------------------------------------------------------------

$\displaystyle
(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\
n=5\\
k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10$

It's 10.

----------------------------------------------------

$\displaystyle
(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\
n=5\\
k=4\\\\
\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90$

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3 years ago