If you know that 3 __<(a/2b)+5<___

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25

answer: a

step-by-step explanation: the answer is a due to the rule of symmetry; that a line of symmetry must divide a figure into 2 pieces that are the exact same, only reflected over a line of symmetry. b and c do not have lines of symmetry and would not divide the figure evenly, thus making the answer a.

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.

proportions are fractions that equal to each other. so if two fractions are the same, they are proportional.

let's check the answers to see if they are proportional.

question a is not proportional because you can't turn 1.5/3 to 3/9. if you try to multiply 1.5/3 by 2 or 3, you wont get 3/9. it's impossible to make the fractions the same.

question b is not proportional because you can't turn 22/33 to 4/6. 33 can't go into 6.

question c is proportional because when you multiply 2 and 3 by 71, you would get 142/213, and those fractions equal to each other

question d is proportional because when you divide 14 and 7 by 2, you would end up with 7/3.5, and those fractions equal to each other.