The expression for 7 times the sum of ( c ) and 2 can be written as ( 7 \times (c + 2) ). This means you first add 2 to ( c ), and then multiply the result by 7. Therefore, the final expression is ( 7(c + 2) ).
5(c+8) or 5c+40
a^2 + b^2=c^2
36
C= 2 times pi and C= pi times diameter C= 2 times pi and C= pi times diameter
-2
7 * (c + 2) = 7 * c + 14
2=w
5(c+8) or 5c+40
a^2 + b^2=c^2
36
int a = 1; int b = 2; int c = a + b; // Sum
The answer will depend on a, b and c. For example, if b = 7 and c = 6 then 60b and 70c are the same number, 420. That has prime factors 2 (twice), 3, 5, 7 which sum to 17. So a would have to be 17/12.
You add 2 fractions with the same denominator [c], so the sum is the sum of the numerators divided by the denominator: a/c + b/c = (a+b)/c
28 1 2 3 4 -------------------| 5 | 6 | 7 -------------------- 8 1 clicks 7 times 2 clicks 6 3 clicks 5 4 c 4 5 c 3 6 c 2 7 c 1 8 c 0 ______ 28
To determine the number of ways to elect 2 student council members from 7 candidates, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 7 ) and ( k = 2 ). So, ( C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 ). Thus, there are 21 different ways to elect 2 members.
To factor this equation... if you have learned before, find 2 numbers that add up to the middle value (-7) and multiply together to have a product of the last value (+10). You can use guess and check for this problem... but if you haven't already found it... the sum of -5 and -2 is -7... and the product of -5 and -2 is +10, so we can use the values -5 and -2.... The solution is (c-5)(c-2) = c^2 - 7c + 10.
C= 2 times pi and C= pi times diameter C= 2 times pi and C= pi times diameter