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Arturiano [62]
3 years ago
10

Rachel rounded 16,473 to the nearest hundred then she rounded it to the nearest thousand what is the final number

Mathematics
2 answers:
castortr0y [4]3 years ago
8 0

17,000 is the answer


deff fn [24]3 years ago
8 0

6 is the thousand digit and we always want to round when we do thousands to the nearest hundred and since she rounded it to the nearest hundred first it should look like this 16,500 then rounded to the nearest thousand is 17,000 we always round up when we have a number more then 5 and round down less than 5.

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Answer:

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Step-by-step explanation:

The volume of the box is:

V = height * length * width

V = x*(66 - 2*x)*(90 - 2*x)

V = (66*x - 2*x^2)*(90 - 2*x)

V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3

V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

subject to:

x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

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From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

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