A,B,C,D are 4 towns. B is 40 kilometeres due east of A. C is 30 kilometeres due north of A. D is 45 kilometeres due south of A.
Worko out the bearing B from C
1 answer:
Answer:
Step-by-step explanation:
The diagram illustrating the scenario is shown in the attached photo.
We would determine angle ACB by applying the tangent trigonometric ratio which is expressed as
Tan θ = opposite side/adjacent side
Taking angle ACB as the reference angle,
Tan C = 40/30 = 1.33
Angle ACB = tan^-1(1.33) = 53.1°
The bearing is calculated with respect to the northern direction. Therefore, the bearing of B from C is
180 - 53.1 = 126.9°
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Upper Tolerance
Remark
The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.
Solution
<u>Upper Limit</u>
Now change the 11/16 into 64. Multiply numerator and denominator or 11/16 by 4
Which results in
With a final result for the fractions of 45/64
So the upper tolerance = 3 45/64
<u>Lower Tolerance</u>
Just follow the same steps as you did for the upper tolerance except you subtract 1/64 like this.
Your answer should be 3 and 43/64
Answer:
2
Step by step explanation:
Given: <u>The average mass of all the students in the class was 35.1 kg. After 1 student left the class, whose mass is 32.3kg, the average mass of the remaining students became 36.3kg, how many students were there in the class room now?</u>
Find out: <u>how many students were there in the class room after one student </u><u>left</u>
Solution: <u>Let the total number of remaining student be= n</u>
<u>Let the total number of remaining student be= nAnd the sum of mass of all the students be = x </u><u>kg</u>
So,
Make x the subject of formula
And
divide both side by n
N = 2.333
so total number of students remaining is 2