<em>Greetings from Brasil...</em>
According to the question of the statement, we can conclude that
PQ = 2B
QR = 2B
PR = base = B
Perimeter = P = 105
P = PQ + QR + PR
105 = 2B + 2B + B
B = 21
<h2>PQ = 2B = 42</h2><h2>QR = 2B = 42</h2>
Answer:
b
Step-by-step explanation:
Answer:
The ships are 66 meters apart.
Step-by-step explanation:
For the sake of convenience, let us label ships A and B
As shown in the figure, the distances to the ships from right triangles.
The distance to the ship A is
and it is given by



And the distance to the ship B is
and is given by



Therefore, the distance
between the ships A and B is

In other words, the ships are 66 meters apart.
Try this option:
rule: logₐ(bc)=logₐb+logₐc;
according to the rule: log₂(4x)=log₂4+log₂x
answer: log₂4+log₂x
Answer and Step-by-step explanation:
The computation of the class width for a frequency is shown below:
Class width is

= 10.86
= 11
Now the class limits for a frequency table
Particulars Lower class limit to upper class limit
First class 20 - 30
Second class 31 - 41
Third class 42 - 52
Fourth class 53 - 63
Fifth class 64 74
Six class 75 - 85
Seventh class 86 - 96