Answer:
i) (0, 2) and (1, 2), ii) (0.333, 1.333) and (1, 2).
Step-by-step explanation:
i) Let be 
, if 
, which is equivalent to the following system of equations:
Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0, 2) and (1, 2)
ii) Let be , if 
, which is equivalent to the following system of equations:
Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0.333, 1.333) and (1, 2)
8*(n+6)=80
distribute
8n+48 = 80
subtract 48 from each side
8n = 32
divide by 8
n=4
Check the picture below.
the constant of proportionality, namely the slope, for the original rectangle will just be 4/2 = 2, and for the dilated version shown that he did, it'll be 20/10 = 2.
part B)
we know the slope is 2, or 2/1 so then
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.
Answer:
They have 48 cards.
Step-by-step explanation:
let s be the number of Shauntay's cards and j be the number of Jessica's cards.
The ratio of Shauntay's cards to Jessica's cards is 5 to 3, then
(1) 
=
After Shauntay gives Jessica 12 of her cards, both girls have the same number of cards, then
(2) s-12 =j+12.
From (2), we get s= j+24. If we put j+24 instead of s in (1) we get:
=
3×j+24×3 = 5×j From this we get;
2×j=72 and j=36
After Shauntay gives Jessica 12 of her cards, Shauntay and Jessica would have 48 cards.
