35 divided by 7 is 5.
a) first find how many numbers are divisible by 5 and 7 separately add them and subtract that are divisible by both 7 and 5 ie 35.the result subtract from 1000 to get the answer to your first question nos. divisible by 5 . last no. div by 5 is 995 .hence we have 995 = 0 + ( n-1) 5 which gives 200 . last no div by 7 is 994 .hence 994 = 0+ ( n-1) 7 which gives n = 143 . nos div by both 5 and 7 .last no div by 35 is 980. 980 = 0+ (n-1) 35 which gives n= 29 hence there are 200+ 143- 29 = 314 numbers div by 5 or 7 a) not div by 5or 7 = 1000- 314 = 686 b)div by 5 and7 =29 c) div by 5 or 7 = 314 d) div by 5 and not by 7 = 200-29 = 171 so answer is 686 Amit singh
The expression equal to 35 can be represented in various ways, such as ( 30 + 5 ), ( 7 \times 5 ), or ( 70 \div 2 ). Each of these expressions simplifies to the value 35.
Sum(not div by 7) = Sum(all) - Sum(div by 7) Now the sum of an AP is Sn = n/2 (first + last) where n is the number of terms Sum(All) = 10000/2 (1 + 10000) = 50005000 Sum(div by 7) = (9996/7)/2 (7 + 9996) = 1428/2 (10003) = 7142142 Sum(not div by 7) = Sum(all) - Sum(div by 7) = 50005000 - 7142142 = 42 862 858
7 is a factor of 35.
7/35 7/7=1 35/7=5 7/35=1/5 1/5=0.2
a) first find how many numbers are divisible by 5 and 7 separately add them and subtract that are divisible by both 7 and 5 ie 35.the result subtract from 1000 to get the answer to your first question nos. divisible by 5 . last no. div by 5 is 995 .hence we have 995 = 0 + ( n-1) 5 which gives 200 . last no div by 7 is 994 .hence 994 = 0+ ( n-1) 7 which gives n = 143 . nos div by both 5 and 7 .last no div by 35 is 980. 980 = 0+ (n-1) 35 which gives n= 29 hence there are 200+ 143- 29 = 314 numbers div by 5 or 7 a) not div by 5or 7 = 1000- 314 = 686 b)div by 5 and7 =29 c) div by 5 or 7 = 314 d) div by 5 and not by 7 = 200-29 = 171 so answer is 686 Amit singh
Sum(not div by 7) = Sum(all) - Sum(div by 7) Now the sum of an AP is Sn = n/2 (first + last) where n is the number of terms Sum(All) = 10000/2 (1 + 10000) = 50005000 Sum(div by 7) = (9996/7)/2 (7 + 9996) = 1428/2 (10003) = 7142142 Sum(not div by 7) = Sum(all) - Sum(div by 7) = 50005000 - 7142142 = 42 862 858
No they have 7 u div:)
35 and 7, respectively.
It's not that hard. If this is what you mean: <div> <div> <h1>Some content</h1> </div> </div>
The element bromine has an atomic number of 35. This means that each atom contains 35 protons, and the requirement for electrical neutrality in an atom means that the atom also contains 35 electrons. A bromide ion contains one more electrons than a bromine atom: 36.
7 is a factor of 35.
35 protons, 36 electrons
7/35 7/7=1 35/7=5 7/35=1/5 1/5=0.2
May be this example will help. Please focus on where ob_implicit_flush(true) and ob_end_flush(); are placed in your code. Version 1:-------------------------------- <?PHP ob_implicit_flush(true); ob_end_flush(); ?> e <div>a<div>a</div> <?PHP sleep(1); ?> <div>b</div> <?PHP sleep(1); ?> c</div> Version 2:------------------------------ <?PHP ob_implicit_flush(true); ob_end_flush(); ?> e<div>a<div>a</div></div> <?PHP sleep(1); ?> <div>b</div> <?PHP sleep(1); ?> c</div> Version 3:------------------------------ <?PHP ob_implicit_flush(true); ob_end_flush(); ?> e<div>a<div>a</div><!--</div>--> <?PHP sleep(1); ?> <div>b</div> <?PHP sleep(1); ?> c</div> ?>
35/7=5 35/5=7. The question is the quotient of 35/5 and 7
x/35 =7 (x/35) *35 = 7 * 35 {multiply both sides by 7} x = 7 * 35 x = 245