Towns p,q and s are shown Q is 35km due east of P Snis 15km due west of p r is 15km due south of p work out the bearing of r fro
m s
1 answer:
Answer:
Step-by-step explanation:
Check attachment for diagram
From the diagram, we can apply Pythagoras theorem to right angle triangle RPS
Then,
Tan θ = opposite / adjacent
Tan θ = 15 / 15
Tan θ = 1
θ = arcTan(1)
θ = 45°
The angle is between the east and south
Then, the bearing of 45°S of east.
The direction is also β
Where
β + θ = 360°
β + 45 = 360°
β = 360 - 45
β = 315°
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<h2>Common ratio = -1/2</h2>
Step-by-step explanation:
term of a Geometric progression is given as
. The first term is given as
.
Any general Geometric progression can be represented using the series .
The first term in such a GP is given by , common ratio by
, and the
term is given by
.
In the given GP,
∴ Common ratio is .
Answer:
And when we apply the limit we got that:
Step-by-step explanation:
Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"
We have the following formula in order to find the sum of cubes:
We can express this formula like this:
And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2
If we operate and we take out the 1/4 as a factor we got this:
We can cancel and we got
We can reorder the terms like this:
We can do some algebra and we got:
We can solve the square and we got:
And when we apply the limit we got that: