The given points are
P = (-4,11)
Q = (-5,8)
The x-component of vector QP is
-4 - (-5) = 1
The y-component of vector QP is
11 - 8 = 3
The vector QP is
(1,3) or

The magnitude of the vector is
√(1² + 3²) = √(10)
Answer:

The magnitude is √(10).
Answer:6.5
Step-by-step explanation:35-15=20, 20-12=8, 8-1.50=6.5
Answer:
y = m x + b equation of a straight line
m m' = -1 condition for perpendicular lines
If y = 4 x - 7 then m = 4 so m' = -.25
Y = -.25 X + A we need to find A
A = Y + .25 * 8 = 2 + 2 = 4
Y = -.25 X + 4
Check:
2 = -.25 * 8 + 4 = -2 + 4 = 2
Answer:
$8882.9
Step-by-step explanation:
A=p(1+(r/n))^nt
Given:P=7000, r=(3÷100%)=0.03 , n=2, t=8
A = 7000(1+(.03/2))^(2×8)
A = 7000 (1+0.015)^16
A = 7000 × 1.015^16
A = $8882.9
A)
Earlier, The length of the angelfish =
inches
Now, the length of angelfish =
inches
We have to determine the grown length of angelfish
=
- 
= 
LCM of '2' and '3' is '6',
= 
=
inch
Therefore, the angelfish has grown by
inch.
B)
We have to determine the increased length of angelfish in feet.
Since
foot
So, 
= 0.069 foot.