a) x+y = 100 b) x-y = 24
a+b:
x+x+y-y=100+24
2x=124
x=62
substitute x=62 into equation a)
y=100-62
y=38
the two numbers are 38 and 62
Divide the two numbers, then multiple by 100
The sum of the even numbers is (26 + 28 + ... + 100); The sum of the odd numbers is (25 + 27 + ... + 99) Their difference is: (26 + 28 + ... + 100) - (25 + 27 + ... + 99) = (26 - 25) + (28 - 27) + ... + (100 - 99) = 1 + 1 + ... + 1 There are (100 - 26) ÷ 2 + 1 = 38 terms above which are all 1; their sum is 38 x 1 = 38. So the difference of the sum of all even numbers and all odd numbers 25-100 is 38.
They are: 100-16 = 84
There are no two prime numbers in which a product of 100 is possible.
To determine what 100 is between 6308 and 5808, we can calculate the difference between the two numbers. The difference is 6308 - 5808 = 500. Therefore, 100 is 100 out of 500, which means it is 20% of the way from 5808 to 6308. Thus, adding 100 to 5808 gives us 5908, which is where 100 falls between the two numbers.
0 and 100.
The numbers 100 and 66, since 100+66=166 and 100-66=34.
The numbers are 53 and 47
The numbers are 51.5 and 48.5
The numbers are 53 and 47
any number A and 100-A would make 100
100 74.4
700
100 & 4
62.5 and 37.5
Divide the two numbers, then multiple by 100
Finding the sum is adding two numbers together. The product comes from multiplying them together. Therefore, the sum of 100 and 100 is 200. The product of 100 and 100 is 10,000. Thus, the difference (which comes from subtracting two numbers) between the two is 9,800.