Someone help me... The following numbers are given: 17, 25, 35, 43, 55, 119, 187. Choose four numbers from the given numbers and write them in the boxes of 2×2 grid so that the numbers in the neighbor boxes are not co-prime and the numbers in non-neighbor boxes are co-prime. (Note that the neighbor boxes are those which share a side).

Step-by-step explanation:

Co-prime numbers are numbers that only have 1 as a common factor.

For example, 35 = 1×5×7, and 39 = 1×3×13.  So 35 and 39 are co-prime.

Write the prime factorization of each number:

17 = 1×17

25 = 1×5²

35 = 1×5×7

43 = 1×43

55 = 1×5×11

119 = 1×7×17

187 = 1×11×17

43 is co-prime with all of these, so we will not use it.

If we start with 35 in the upper left, and 187 in the lower right, then we can also rule out 17 and 25, since these are co-prime with either 35 or 187.

So that leaves 55 and 119 as the other two numbers.  They can go in any order, as long as they are diagonal from each other.

yes

step-by-step explanation:

yes

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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