A student walk 60m on a bearing
1 answer:
Answer:
d = 234.6 m
Step-by-step explanation:
You can consider a system of coordinates with its origin at the beginning of the walk of the student.
When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:
she is at the coordinates (52.97 , 28.16)m.
Next, when she walks 180m to the east, her coordinates are:
(52.97+180 , 28.16)m = (232.97 , 28.16)m
To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:
The displacement of the student on her complete trajectory was of 234.6m
You might be interested in
Answer:
- Inverse Variation
- Equation:
Step-by-step explanation:
Given
X-> 2 | 4 | 8 | 12
Y-> 6 | 3 | 3/2 | 1
Required
- State the type of variation the table represents
- Determine the equation to model the data
To check for type of variation; we'll make use of trial by error method.
To start with; We'll check for direct variation.
This is done using the following expression;
where k is the constant of variation
Make k the subject of formula
When Y = 6, X= 2
When Y = 3, X = 4
There's no need to check further as both values of k are not equal
To check for inverse variation;
Make k the subject of formula
When Y = 6, X= 2
When Y = 3, X = 4
When Y = 3/2; X = 8
When Y = 1, X = 12
Note that for all values of X and Y, K remains constant;
Hence, the table represents an inverse direction
To determine the equation;
We make use of
Substitute 12 for k
So, the equation becomes
Reorder
Hence, the equation is