Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
Your question answers itself. You solve by graphing.
I like to subtract one side of the equation from the other, so the solutions are where the graph crosses the x-axis (the resulting function value is zero).
It can be useful to find the "turning point" of each absolute value expression (where its value is zero) and graph that and some points on either side.
Mass error is due to temperature which is a random error. I believe.
Jesus is the lord above all
1 and 18
1 × 18 = 8
1+ 18 = 19
