Answer:
155.3 m
Step-by-step explanation:
<u>We have:</u>
θ: is the angle measured from the vertical = 75 °
v = is the speed of the submarine = 10 m/s
t = 1 min
The deep of the front end of the submarine at the end of a 1-minute drive is given by the following ratio:
<u>Where:</u>
d: is the distance traveled by the submarine in 1 minute
x: is the deep to find
The distance, d, is:
Now, the deep is:
Therefore, the deep of the front end of the submarine at the end of a 1-minute drive is 155.3 m.
I hope it helps you!
1 km = 0.621 mi
5 km = 3.105
All you need is to do is 5 x 0.621
Answer:
x= 0 , 
, 
Step-by-step explanation:
Given, equation is ![\sqrt[3]{15x-1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B15x-1%7D)
+ ![\sqrt[3]{13x+1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B13x%2B1%7D)
= ![4\sqrt[3]{x}](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7Bx%7D)
. →→→ (1)
Now, by cubing the equation on both sides, we get
( + )³ = ()³
⇒ (15x-1) + (13x+1) + 3×× ( + ) = 64 x.
⇒ 28x + 3×× () = 64x.
(since from (1), + = )
⇒ 12× × (![\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D)
)= 36x.
⇒ 3x = ![\sqrt[3]{(15x-1)(13+1)(x)}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%2815x-1%29%2813%2B1%29%28x%29%7D)
.
Now, once again cubing on both sides, we get
(3x)³ = ()³.
⇒ 27x³ = (15x-1)(13x+1)(x).
⇒ 27x³ = 195x³ + 2x² - x
⇒ 168x³ + 2x² - x = 0
⇒ x(168x² + 2x -1) = 0
⇒ by, solving the equation we get ,
x = 0 ; x = ; x =
therefore, solution is x= 0 , ,
The answer would be saturday because you must continue adding five
Answer:180 I think
Step-by-step explanation:
