Answer: PN= 20 in.
Step-by-step explanation:
Given: In triangles PMN and QSR :
Length of P N ≅ Length of RQ (1) [ '≅' → Sign of congruence]
Length of QS ≅ Length of PM
Length of MN ≅ Length of RS
By SSS congruence rule
ΔPMN ≅ ΔQSR
Since Length of QR =20 in. [Given]
From (1)
Length of PN= 20 in. [Congruent means exactly equal]
It equals 14,000 is the correct answer
Answer:
Step-by-step explanation:
We have been given an expression 
. We are supposed to combine like terms for our given expression.
We can see that 
are like terms as both of these terms have x variable.
Similarly, 
are like terms as both are constant.
Upon combining like terms, we will get:
Therefore, our required expression would be .
Perform division as shown here
You should get 7 1/2
A Canadian postal code looks like this:
K1A 3B1 .
So you have: letter - digit - letter - digit - letter - digit .
The question doesn't say anything about restrictions on
which letters can be used, or restrictions on repeating letters
or digits within one postal code. So as far as we know, each
letter can be any one of 26, and each digit can be any one of 10.
The total number of possibilities would be
(26·10·26) · (10·26·10) = 17,576,000 .
In the real world, though, (or at least in Canada), Postal codes
don't include the letters D, F, I, O, Q or U, and the
first letter
does not use W or Z. When you work it out with these restrictions,
it means there's a theoretical limit of 7.2 million postal codes.
The practical limit is a bit lower, as Canada Post reserves some
codes for special functions, such as for test or promotional purposes.
One example is the code H0H 0H0 for Santa Claus ! Other special
codes are for sorting mail
bound for destinations outside Canada.
At the present time, there are a little over 830,000 active postal codes.
That's about 12% of the total possibilities, so there are still plenty of codes
left for expansion.
