Answer:
14 and 16.
Step-by-step explanation:
We know that two consecutive even numbers is 30.
Let's let n be <em>any number</em>.
Then the first <em>even</em> number must be 2n.
This is because n can be any number, either even or odd. However, if we multiply it by 2, this <em>ensures</em> that 2n is even. Remember that any number multiplied by 2 yields an even number.
Therefore, the consecutive even number will be (2n+2). Not (2n+1), because that gives an odd number.
We know that they sum to 30. So, we can write the following equation.

And we solve from there:

So, the value of n is 7.
Therefore, the first even number is 7(2)=14.
And the consecutive even number is 16.
Edit: Typo
19-15 = C Number of students over the course of a year!
19-15 gives us a total of 4 students over the course of the year!
So the value of C is 4 students
Actually I think it may be C
Answer:
21 matches
Step-by-step explanation:
This is a combination problem where we have to select two member out of 7 members
So we want calculation of 7C2
7C2 = 7!/[(2!)(7-2)!]
(7*6*5*4*3*2*1)/[(2*1) *(5*4*3*2*1]
(7*6)/2 ~ Note everything that easily cancels here!
42/2
21 matches