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DedPeter [7]
2 years ago
15

Truncate the decimal expansion of 15 to find the best approximation. A) 2.4 < 15 < 2.5 B) 2.46 < 15 < 2.47 C) 2.46 &

lt; 15 < 2.50 D) 2.47 < 15 < 2.50
Mathematics
2 answers:
leva [86]2 years ago
8 0

This is not the complete question.

In this we have to truncate the decimal expansion of ^3\sqrt{15} to find the best approximation. The options are:

A) 2.4 <^3\sqrt{15} < 2.5

B) 2.46 < ^3\sqrt{15} < 2.47

C) 2.46 <^3\sqrt{15} < 2.50

D) 2.47 < ^3\sqrt{15} < 2.50

Solution: Consider the cube root of 15,

By using the calculator, we will calculate its cube root.

^3\sqrt{15}=2.4662

Now, we have to find the option which approximates it in the best way.

In option A. It is given between 2.4 and 2.5 which is not the best approximation as the decimal expansion is 2.46 is between 2.46 and 2.47.

In Option B: It is given between 2.46 and 2.47 which is the best approximation as the decimal expansion is between 2.46 and 2.47.

In Option C: It is given between 2.46 and 2.50 which is not the best approximation as the decimal expansion is 2.46 is between 2.46 and 2.47.

In Option D: It is given between 2.47 and 2.50 which is not the best approximation as the decimal expansion is 2.46 is between 2.46 and 2.47.

Hence, The decimal expansion of ^3\sqrt{15} is between 2.46 and 2.47.

So, we get 2.46 < ^3\sqrt{15} < 2.47

Therefore, Option B is the correct answer.

chubhunter [2.5K]2 years ago
7 0
∛15 ≈ 2.466212...

The best choice appears to be ...
  B) 2.46 < ∛15 < 2.47
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V (max)  = 2 ft³

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Step-by-step explanation:  See annex ( two different cubes)

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And  the height   is x  Then volume of the cube as a function of x is:

V(x)  =  ( 3 - 2x )² *x   or       V(x)  =    ( 9 + 4x² - 12x )*x

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Taking derivatives on both sides of the equation

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Second degree equation solving for x

x₁,₂ =  [ 24 ± √( 576) - 432  /24

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Of these two values we have to dismiss x₁  because if  x = 1.5 we don´t have a cube ( 0 height )

Then we take x  = 0,5  feet

And

V (max)  =  (2)²*0,5   =  4*0,5

V (max)  = 2 ft³

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