Vectors: A boat moves through the ocean on a foggy day. The boat is pointing North and can typically move at a speed of 12.8 m/s in calm waters. The boat travels through water where there is a 5.9 m/s current flowing to the East. What is the speed of the boat relative to a stationary observer on land? __m/s Part 2 The boat turns to face directly West. Now how fast is the boat moving relative to land? __m/s

1.) 14.09 m/s

2.) 6.9 m/s

Step-by-step explanation:

Given that the boat is pointing North and can typically move at a speed of 12.8 m/s in calm waters. The boat travels through water where there is a 5.9 m/s current flowing to the East.

The speed of the boat relative to a stationary observer on land will be achieved by using pythagorean theorem

V^2 = 12.8^2 + 5.9^2

V^2 = 198.65

Square root both sides

V = 14.09 m/s

2.) If the boat turns to face directly West.

Since the boat can move at speed 12.8 m/s in a calm water, the speed of the boat if it face west will be

12.8 - 5.9 = 6.9 m/s

Therefore, the boat is moving relative to land at 6.9m/s

step-by-step explanation:

it's is b

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72