Writing an Equation
The depth of the water at the end of a pier changes periodically along with the movement of tides. On a
particular day, low tides occur at 12:00 am and 12:30 pm, with a depth of 2.5 m, while high tides occur at
6:15 am and 6:45 pm, with a depth of 5.5 m. Let t = 0 be 12:00 am.
Write a cosine model, d = acos(bt) + k, for the depth as a function of time.
meters.
This amplitude is
a =
DONE

statement #3: proof- ang is congruent to ang ecd : angles are congruent.  

step-by-step explanation:

1: the missing angles in statement #3  

a.) acb  

reason- the reason why the answer is acb is because angle ecd and angle acb are both the same size *that's what the word congruent means this is really simple.  

2: the missing word in reason #3  

c.) vertical  

reason- the reason why the answer is vertical is because the angles in that diagram on figure 2 is slanted, which is vertical, and the e and a both lay on vertical lines. along with the other's, but e and a are the 2 main points you want to focus on. and c is kind of like your starting point, because c is for both angles, not just one.  

i hope that was and understanding.

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.