Suppose the depth of the tide in a certain harbor can be modeled by y = 23 - 5 cos pi/6 t, where y is the water depth in feet and t is the time in hours. Consider a day in which t = 0 represents 12:00 midnight. For that day, when are high tide and low tide and what is the depth of each? a.
high tide at 6:00 a. m. and 6:00 p. m., depth 28 ft
low tide at 12:00 noon and 12:00 midnight, depth 18 ft
b.
high tide at 6:00 a. m. and 6:00 p. m., depth 23 ft
low tide at 12:00 noon and 12:00 midnight, depth 5 ft
c.
high tide at 12:00 noon and 12:00 midnight, depth 28 ft
low tide at 6:00 a. m. and 6:00 p. m., depth 18 ft
d.
high tide at 12:00 noon, depth 23 ft
low tide at 6:00 a. m., depth 5 ft

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841