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miss Akunina [59]
2 years ago
11

(5√3-√27)^3 how do you prove that this is an integer?

Mathematics
2 answers:
mr Goodwill [35]2 years ago
5 0

Keys:

  • \left(a\cdot \:b\right)^n=a^nb^n

Step-by-step explanation:

\left(5\sqrt{3}-\sqrt{27}\right)^3\\\5\sqrt{3}-\sqrt{27}^3=\left(2\sqrt{3}\right)^3\\= 2\sqrt{3}\\\left(2\sqrt{3}\right)^3=2^3\left(\sqrt{3}\right)^3\\=2^3\left(\sqrt{3}\right)^3\\2^3=8\\=8\left(\sqrt{3}\right)^3\\=8\cdot \:3\sqrt{3}\\8\cdot \:3=24\\=24\sqrt{3}

kirill [66]2 years ago
4 0

Answer:

It is not an integer.

Step-by-step explanation:

(5\sqrt{3} -\sqrt{27} )^3\\\\=(\sqrt{3} *(5-3))^3\\\\=8*3*\sqrt{3} \\\\=24\sqrt{3} \\

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Step-by-step explanation:

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When a number goes over the equal sign= positive + changes to negative, vice versa.

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igomit [66]

Answer:

\large\boxed{x=-\dfrac{7-\sqrt{97}}{4}\ \vee\ x=-\dfrac{7+\sqrt{97}}{4}}

Step-by-step explanation:

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The function h(t) = -16t2 + 48t + 28 models the height (in feet) of a ball where t is the time (in seconds).
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<em>[using calculus] </em>When the function h(t) reaches its maximum value, its first derivative will be equal to zero (the first derivative represents velocity of the ball, which is instantaneously zero). We have h'(t) = -32t + 48, which equals zero when t = 3/2 = 1.5. The ball therefore reaches its maximum height when t = 1.5. To find the maximum height, we need to find h(1.5), which is 64 feet.

<em>[without calculus] </em>This is a quadratic function, so its maximum value will occur at its vertex. The formula for the x-coordinate of the vertex is -b/2a, so the maximum value occurs when t = -48/(2*16), which is 1.5. The maximum height is h(1.5), which is 64 feet.

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