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)
Answer:-12^8
Step-by-step explanation:
Answer:
307 paces
Step-by-step explanation:
First, lets convert Robert's pace to metres. (cm > m)
75cm = (75/100)m
= 0.75m
Number of Full Paces = Length of Bridge / 1 Full Pace
=

See , we must have full paces, and in this case we are finding the minimum number of paces, so we will be looking at the lowest possible whole number, which in this case, is 307 paces.
find the prime factors in each term to find the GCF
answer is 6x^2