The English alphabet contains 26 letters (a, b, c, ...y, z).
The digits from 0 to 9 are a total of 10.
A keycode contains 2 letters, and 3 numbers, for example:
AB 598; MM 139; NT 498; ...
So there are 26 possible choices for the first letter, which can combined with any of the 26 possible choices for the second letter, so there are a total of
26*26=676 possible pairs of letters.
Similarly, the 10 possible choices for the first number, which can be combined with the 10 possible choices for the second number, and the 10 possible choices for the third number make a total of :
10*10*10=1,000 possible triples of numbers.
Any of the 676 possible pairs of letters can be combined with any of the possible 1,000 triples of numbers. This makes a total of
676*1,000=676,000 keycodes.
Answer: 676,000
1. d
x-intercept (y=0) is -4
y-intercept (x=0) is 8
2. d
x-intercept (y=0) is 16
y-intercept (x=0) is 20
3. b
y=1/6x+4
multiplied by 6
6y=x+24
-x+6y=24
4. a 4.75k+2.25p=22
So to find the width (w)
26=2(7+w). W=6
The numbers are 3.5 and -10.5
Step-by-step explanation:
- Step 1: Sum of two numbers = -7. Let one number be x. Then, the other number = -7 - x. Difference = 14
⇒ x - (-7 - x) = 14
⇒ x + 7 + x = 14
⇒ 2x = 7
∴ x = 7/2 = 3.5
-7 - x = -7 - 3.5 = -10.5
