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3 years ago

Find three consecutive positive integers such that the sum of their squares is 2354. What is the largest integer?

Mathematics

1 answer:

Ilya [14]3 years ago

4 0

Answer:

The largest integer is 29.

Step-by-step explanation:

Let the consecutive positive integers are x, x+1 and x+2.

The square of sum of squares of three consecutive positive integers is 2354 such that,

$x^2+(x+1)^2+(x+2)^2=2354\\\\x^2+x^2+2x+1+x^2+4+4x=2354\\\\3x^2+6x+5=2354\\\\3x^2+6x- 2349=0$

It is a quadratic equation whose solution is given by :

x = 27 and x = -29

First positive integer = 27

Second positive integer = 27+1 = 28

Third positive integer = 27+2 = 29

Hence, the largest integer is 29.

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4 years ago

How do you find the answer?

Jet001 [13]

1)Sin(29)=x/(30cm)
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3 years ago

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3 years ago

An army contingent of 104 members is to march behind an army band of 96 members in a parade.

Trava [24]

You can dispose a number $x$ of elements in a matrix-like formation with $n\times m$ shape if and only if $n$ and $m$ both divide $x$ , and also $nm=x$ .

So, we need to find the greatest common divisor between $104$ and $96$ , so that we can use that divisor as the number of columns, and then.

To do so, we need to find the prime factorization of the two numbers:

$104 = 2^3\times 13$

$96 = 2^5 \times 3$

So, the two numbers share only one prime in their factorization, namely $2$ , but we can't take "too many" of them: $104$ has "three two's" inside, while $96$ has "five two's" inside. So, we can take at most "three two's" to make sure that it is a common divisor. As for the other primes, we can't include $3$ nor $13$ , because it's not a shared prime.