If Jane has 23 cats and I have 2 cats, and then Jane gives me 5 cats, how many more cats does Jane have than I? Rhonda has 12 ma
1 answer:
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Answer:
The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.
Step-by-step explanation:
The question is incomplete. The complete question should be
The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?
Given the initial bearing of a lighthouse from the ship is N 37° E. So, is 37°. We can see from the diagram that
would be
143°.
Also, the new bearing is N 25°E. So, would be 25°.
Now we can find . As the sum of the internal angle of a triangle is 180°.
Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.
And let us assume the distance between the lighthouse and the ship at N 25°E is
We can apply the sine rule now.
So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.
Answer:
1/32
Step-by-step explanation:
This could be written as
1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/32, or 0.03125.
Answer:
2063.9 kg
Step-by-step explanation:
Given,
<u>1 kg = 2.2046 pounds</u>
This can also be written as:
<u>1 pound = 1/2.2046 kg</u>
We have to calculate the value of 4550 pounds in kg up to nearest 10th
Thus,
Solving the above equation, we get:
4550 pounds = 2063.8664 kg
Rounding the above result to nearest tenth as:
<u>4550 pounds = 2063.9 kg</u>