How many positive integers less than 1000 a) aredivisibleby7? b) are divisible by 7 but not by 11? c) are divisible by both 7 and 11? d) are divisible by either 7 or 11? e) are divisible by exactly one of 7 and 11? f) aredivisiblebyneither7nor11? g) havedistinctdigits?

a.142 b.130 c.12 d.220 e.208 f.779 g.720

Step-by-step explanation:

How many positive integers less than 1000 (from 1 to 999)

a) are divisible by 7?

999/7=142,71

then there are 142 integers that are divisible by 7

7*1=7

7*2=14

7*3=21

...

7*141=987

7*142=994

b) are divisible by 7 but not by 11?

as 7 and 11 are prime numbers

7*11=77

999/77=12.97

142-12=130

c) are divisible by both 7 and 11?

999/77=12.97

then there are 12 integers that are divisible by both 11 and 7 below 1000

d) are divisible by either 7 or 11?

999/7=142.71

999/11=90.81

then the numbers that are divisible either by 7 or 11 is

142+90-12(the numbers that are counted in both grups 'divisible by 7' and 'divisible by 11')

220

e) are divisible by exactly one of 7 and 11?

[142('divisible by 7')-12('divisible by both')]+[90('divisible by 11')-12('divisible by both')]

208

f) are divisible by neither 7 nor 11?

999('all the positive integers below 1000)-220('are divisible by either 7 or 11')

779

g) have distinct digits?

For evey digit x from 0 to 9

0xx,1xx,2xx,3xx,4xx,5xx,6xx,7xx,8xx,9xx 10 options

xx0,xx1,xx2,xx3,xx4,xx5,xx6,xx7,xx8,xx9 10 options

x0x,x1x,x2x,x3x,x4x,x5x,x6x,x7x,x8x,x9x 10 options

and xxx one option

for each of the first three rows we have to sustract the option in wich the three digits are the same digit.

for the 0 we have to sustract the 000 of the fourth row as it is not a positive number.

10 (different digits)*(9+9+9+1)(different combinations for each number)-1(the 000)=279

999-279=720

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1