Mathematicians use "..." to mean that a pattern continues: it means to fill in everything in-between.
Let A represent 1+3+5+ ... + 99, the sum of all the odd
numbers less than 100.
Let B represent 2+4+6... + 98, the sum of all the even
numbers less than 100.
Which is bigger, A or B, and by how much? Do this without adding
them up the long way. A full solution to this challenge problem
means that you must explain why your answer is correct - you
can't just give a number, you have to prove it!

Step-by-step explanation:

We khow that A is the sum of all odd numbers that are less than 100

So A= 1+3+5+...+99

We can calculate this sum without adding all this numbers one by one

A = the number of terms *( first term + last one ) over 2

To get the number of terms we substract the first term from the last term then we add one

We khow that odd numbers are wiritten this way : 2*n +1 where n is an integer

So 99= 2*49+1

1= 2*0+1

So the number of terms is : 49-0+1 =50

So A= 50*(1+99)/2 = 2500

Foloowing the same method we get :

B= (49-1+1)*(2+98)/2= 2450

A>B

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1