Find the sum of all natural numbers between 25 and 210 which are either divisible by 3 or divisible by 4? Please let me know ​I will mark you Brainliest

Let S be the sum of the integers 25-210:

S = 25 + 26 + 27 + … + 208 + 209 + 210

Let S₃, S₄, and S₁₂ denote the sums of the integers in S that are multiples of 3, 4, or 12, respectively. We'll also count how many terms each sum involves; it'll be useful later.

S₃ = 27 + 30 + 33 + … + 204 + 207 + 210 … … … (62 terms)

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

S₄ = 28 + 32 + 36 + … + 200 + 204 + 208 … … … (46 terms)

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₁₂ = 36 + 48 + 60 + … + 180 + 192 + 204 … … … (15 terms)

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

Let's look at S₃ :

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

By reversing the order of the sum, we get

S₃* = 3 (70 + 69 + 68 + … + 11 + 10 + 9)

Of course S₃ = S₃*, I'm just calling it something else temporarily. Notice that every term in the same position of either sum adds up the same number.

9 + 70 = 79

10 + 69 = 79

11 + 68 = 79

and so on. Then

S₃ + S₃* = 3 (79 + 79 + 79 + … + 79 + 79 + 79)

or

2S₃ = 3 × 62 × 79   ==>   S₃ = 7,347

We can compute the other two sums in the same way.

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₄* = 4 (52 + 51 + 50 + … + 9 + 8 + 7)

==>   2S₄ = 4 × 46 × 59  ==>   S₄ = 5,428

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

S₁₂* = 12 (17 + 16 +15 + … + 5 + 4 + 3)

==>   2S₁₂ = 12 × 15 × 20   ==>   S₁₂ = 1,800

Then the sum you want is

S₃ + S₄ - S₁₂ = 10,975

We subtract S₁₂ because each of its terms is counted twice (once in S₃ and again in S₄).

they would have to work 32 hours.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.