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2 years ago

The sum of any two different odd numbers is divisible by 4.

Mathematics

2 answers:

bagirrra123 [75]2 years ago

4 0

Answer:

False

Step-by-step explanation:

counterexample: 3 + 7 = 10 . . . . not divisible by 4

The statement is FALSE.

An odd number can be represented by ...

2n+1 . . . . where n is any integer

Then a different odd number can be represented by ...

2m +1 . . . . where m is any integer

The sum of these numbers will be ...

(2n +1) +(2m +1) = 2n +2m +2 = 2(n +m +1)

This will be divisible by 2 for any integers m and n, but will only be divisible by 4 if exactly one of the integers m and n is odd.

jeyben [28]2 years ago

3 0

Step-by-step explanation:

are 1 and 3, then their sum is 4 which is divisible by 4.

are 3 and 5, their sum is 8 which is divisible by 4.

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Terrell loves to listen to music, so he buys a subscription to a music-streaming service. He pays $4.99 each month. How much doe

Ilya [14]

Answer:

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Step-by-step explanation:

Given

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3 years ago

The terms in the sequence decrease by the same amount each time:

luda_lava [24]

Answer:the nth term of the sequence, Tn= -3n +12

Step-by-step explanation:

The nth term of an arithmetic progression is given as

Tn= a+ (n-1) d

where a= first term

and d = common difference

In this sequence, 9, 6 ,3 ,0, -3, -6 we can see that the number is decreasing by 3

The first term, a= 9

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using our formulae

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7 0

3 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

noname [10]

$L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)$

$L_x=10+10\lambda x=0\implies1+\lambda x=0$

$L_y=10+10\lambda y=0\implies1+\lambda y=0$

$L_z=2+4\lambda z=0\implies1+2\lambda z=0$

$L_\lambda=5x^2+5y^2+2z^2-42=0$

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$z^2=1\implies z=\pm1\implies x=y=\pm2$

There are two critical points, at which we have

$f\left(2,2,1\right)=42\text{ (a maximum value)}$

$f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}$

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3 years ago

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