A sports teacher had 1025 sweets. After distributing these equally amongst all the participants in a sports meet, he had just 1 sweet left with him. Which of these could be the number of sweets given away to each participant?

Step-by-step explanation:

Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).

As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):

1) 512 => 2 partecipants

2) 256 => 4 partecipants

3) 128 => 8 partecipants

4) 64 => 16 partecipants

5) 32 => 32 partecipants

6) 16 => 64 partecipants

7) 8 => 128 partecipants

9) 4 => 256 partecipants

10) 2 => 512 partecipants

The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.

Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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