Let's call the 10 runners A B C D E F G H I J
ABC
ABD
ABE
ABF
ABG
ABH
ABI
ABJ
ACB
ACD
ACE
ACF
ACG
ACH
ACI
ACJ
ADB
ADC
ADE
ADF
ADG
ADH
ADI
ADJ
AEB
AEC
AED
AEF
AEG
AEH
AEI
AEJ
AFB
AFC
AFD
AFE
AFG
AFH
AFI
AFJ
AGB
AGC
AGD
AGE
AGF
AGH
AGI
AGJ
AHB
AHC
AHD
AHE
AHF
AHG
AHI
AHJ
AIB
AIC
AID
AIE
AIF
AIG
AIH
AIJ
AJB
AJC
AJD
AJE
AJF
AJG
AJH
AJI
The above is 60 different ways, but now B C D E F G H I J have to come first, and the proses starts again, until we have covered them all. I'm not going to write them all out but in theory, there is 540 different ways. 9*60=540.
Answer:
Step-by-step explanation:
1. sinФ = cos 25
25 ° is in the between 0 and 90°
therefore it can simply represent
cosФ= sin (90-Ф) = sin (90 - 25) = sin 65
2. sin(Ф/3 + 10) = cos Ф
cos Ф = sin (90 -Ф)
sin(Ф/3 + 10) = cos Ф
sin(Ф/3 + 10) = cos Ф = sin(90-Ф)
Ф/3+10=90-Ф
10Ф+3/30 = 90-Ф
10Ф+3 = 30(90-Ф)
10Ф+3 = 2700-30Ф
10Ф+30Ф=2700-3
40Ф = 2697
Ф = 2697 / 40 = 67.425 ≅ 67.4°
How many 6-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, if repetitions of digits are allowed?
sveta [45]
There are 6 digits. Each digit can take ten different numbers except for the first digit since it cannot be zero.
So:
9 x 10 x 10 x 10 x 10 x 10
900000 numbers.
Another way of thinking about this is to just count up to 999,999. Obviously there are 999,999 different numbers here. But since our number has to have 6 digits in them, we have to delete 99,999 numbers. Thus there are 900,000 different numbers.
Answer:
a+0= a
Step-by-step explanation:
we know that
The <u>additive identity</u> property says that if you add a real number to zero or add zero to a real number, then you get the same real number back
so
Let
a -----> a real number
a+0=0+a=a
therefore
a+0= a
Can you show your original question
