Answer: 2077
There are 31 terms.
S₃₁ = 31(a₁+a₃₁)/2 = 31(7+127)/2 = 2077
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:
AOD=180 so COD=48
Since point C and D are equidistance to center point(O) OD and OC are congruent.
That also means OCD=ODC.
180-48=OCD+ODC
ODC=OCD
x= 66
Answer:
if you are lookng for C it is 17
Step-by-step explanation:
Did this a couple years ago, im kinda lost xD . sorry
