Answer:
We can use seven letters and numbers.
I am assuming that any numeral in the range 0..9 or any letter from the English alphabet A..Z can appear in any position, with no blank spaces allowed and no restrictions on repetition. I am also assuming that plates with fewer than seven letters and numbers are not allowed.
So, for example A879BX8 is acceptable, so are 5555555 and ABCDEFG, but not A.123.ZX or…..7A, where the dot represents a space.
I am also assuming that you can only use upper case letters.
With these restrictions, there are 36 possibilities for each space and the total number of valid number plates would be 36^7 = 78,364,164,096, let's say about 78 billion.
It is estimated that there are about 1.3 billion cars, trucks and buses in the road today. This number plate system therefore allows more than enough unique license plates. I'd even hazard a guess that it might be more than enough for every road vehicle that has ever been built or ever will be.
In practice there would be other restrictions, for example only letters in some positions and only numbers in others. There'd still be plenty to go around.
Step-by-step explanation:
Answer:
Step-by-step explanation:
f * g = (x^2 + 3x - 4) (x+4)
open bracket
x((x^2 + 3x - 4) + 4 (x^2 + 3x - 4)
x³ +3x²-4x+x²+12x-16
x³+3x²+x²-4x+12x-16
x³+4x²+8x-16 (domain is all real numbers.
f/g = (x^2 + 3x - 4)/(x+4)
factorising (x^2 + 3x - 4)
x²+4x-x_4
x(x+4) -1 (x+4)
(x+4)(x-1)
f/g = (x^2 + 3x - 4)/(x+4) =(x+4)(x-1)/(x+4) = (x-1)
Before factorisation, this was a rational function so the domain is all real numbers excluding any value that would make the denominator equal zero.
Hence I got x - 1, and x cannot equal -4
So the domain is just all real numbers without -4
Answer:
yes they are
Step-by-step explanation:
because the metric system is the same
Answer:
B
Step-by-step explanation:
Both of them have the same variable and are both raised to the 5th power (Like the example in the last sentence of the text)
