Earth's oceans have an average depth of 3800 m, a total area of 3.63 × 108 km2, and an average concentration of dissolved gold of 5.8 × 10−9 g/l. how many grams of gold are in the oceans?

grams of gold=8x10^12 g

Step-by-step explanation:

first we calculate the volume of the ocean this is achieved by multiplying the area by the depth

A=3.63x10^8  km^2=3.63x10^14   m^2

V=AxL

V=3.63x10^14 x 3800=1.3797x10^18m^3=1.3797x10^21L

then we multiply the volume in liters by the concentration of gold in the ocean per liter

grams of gold=1.3797x10^21   x    5.8x10^-9=8x10^12 g

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

a: inscribed angles

step-by-step explanation:

i reviewed this question with a peer and teacher, seems to be correct, me later