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alexira [117]
2 years ago
8

If a = 7 - 4√3, find the value of a + 1/a​

Mathematics
1 answer:
irga5000 [103]2 years ago
5 0

Answer:

Step-by-step explanation:

To calculate a+1/a, we first need to calculate for a.

a = 7 - 4 * \sqrt[2]{3}

square root of 3 = 1.73

a = 7 - 4 * 1.73

a = 7 - 6.9

a = 0.1

0.1 + 1 / 0.1 = 0.1 + 10 = 10.1

Now, in case 4 wasn't being multiplied with the square root of 3, and instead, it was four root of 3, I am gonna do the calculations again:

a = 7 - \sqrt[4]{3}

a = 7 - 1.31

a = 5.69

5.69 + 1 / 5.69 = 5.69 + 0.17 = 5.86

Hope I Helped!

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What is the common denominator of (5/x^2-4) - (2/x+2) in the complex fraction (2/x-2) - (3/x^2-4)/(5/x^2-4) - (2/x+2)
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9514 1404 393

Answer:

  • common denominator: (x² -4)
  • simplified complex fraction: (2x +1)/(9 -2x)

Step-by-step explanation:

It is helpful to remember the factoring of the difference of squares:

  a² -b² = (a -b)(a +b)

__

Your denominator of (x² -4) factors as (x -2)(x +2). You will note that one of these factors is the same as the denominator in the other fraction.

It looks like you want to simplify ...

  \dfrac{\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}\right)}{\left(\dfrac{5}{x^2-4}-\dfrac{2}{x+2}\right)}=\dfrac{\left(\dfrac{2(x+2)}{(x-2)(x+2)}-\dfrac{3}{(x-2)(x+2)}\right)}{\left(\dfrac{5}{(x-2)(x+2)}-\dfrac{2(x-2)}{(x-2)(x+2)}\right)}\\\\=\dfrac{2(x+2)-3}{5-2(x-2)}=\boxed{\dfrac{2x+1}{9-2x}}

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