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3 years ago

What are two consecutive numbers whose cube differs by 271

Mathematics

2 answers:

valentina_108 [34]3 years ago

7 0

Answer:

Two consecutive numbers cubes differ by 271. Let the two consecutive numbers be x and x + 1. The difference of the two consecutive numbers is 271. ( x + 1 ) ^3 - x ^3 = 271. 1 + 3 x ^ 2 + 3 x - 271 = 0.

Step-by-step explanation:

lesya692 [45]3 years ago

4 0

9514 1404 393

Answer:

9, 10 or -10, -9

Step-by-step explanation:

Let x represent the smaller of the two numbers. Then x+1 is the larger, and their difference is ...

(x +1)³ -x³ = 271

x³ +3x² +3x +1 -x³ = 271

3x² +3x -270 = 0 . . . . . . simplify

x² +x -90 = 0 . . . . . . . . divide by 3

(x +10)(x -9) = 0 . . . . . factor

The two solutions are x = -10, x = 9.

The numbers could be {-10, -9} or {9, 10}.

_____

<em>Additional comment</em>

Solutions to a factored polynomial are the values of the variable that make the factors zero. The product will only be zero if a factor is zero.

The factor (x -a) is zero when x=a.

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A town has a population of 14000 and grows at 3% every year. What will be the population after 6 years, to the nearest whole num

soldi70 [24.7K]

This problem resembles one of a compound interest type, in which we would use the following formula:

$M = C . (1 + j)^{n}$

Where:

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Using this same formula to solve our problem, we have:

C is the initial population

j is the annual growth rate

n is the number of years

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OK. Let's do it...

C = 14000

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Making the due substitutions and performing the calculations:

M = C . $(1 + j)^{n}$

M = 14000 . $(1 + .03)^{6}$

M = 14000 . $(1 .03)^{6}$

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Response: 16,717

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