44. the decimal expansions of e and √5 are given below.
e ≈ 2.71828182…
√5 ≈ 2.23606797…
a. find an approximation of √5 +e to three decimal places. do not use a calculator.

√5 +e = 4.953

Step-by-step explanation:

given data

e ≈ 2.71828182…

√5 ≈ 2.23606797

to find out

√5 +e

solution

√5 +e = 2.71828182 + 2.23606797

√5 +e = 2.718282 + 2.236068

√5 +e = 4.95435

√5 +e = 4.953

it would not be increasing by nothing that i can see probably by 25 percent that is all i can sat\y for now if i have any thing else to say i will edit this.

step-by-step explanation:

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.