How many pairs of whole numbers have a sum of 40

1 answer:

Alex3 years ago

5 0

For the writer, there are 20
1. 1 + 39 = 40 
2. 2 + 38 = 40
3. 3 + 37 = 40
4. 4 + 36 = 40 
5. 5 + 35 = 40 
6. 6 + 34 = 40
7. 7 + 33 = 40
8. 8 + 32 = 40 
9. 9 + 31 = 40
10. 10 + 30 = 40 
11. 11 + 29 = 40 
12. 12 + 28 = 40
13. 13 + 27 = 40 
14. 14 + 26 = 40 
15. 15 + 25 = 40 
16. 16 + 24 = 40 
17. 17 + 23 = 40 
18. 18 + 22 = 40 
19. 19 + 21 = 40 
20. 20 + 20 = 40

You might be interested in

I neeeeeeeeeeeeedddddd itttttttttt

LekaFEV [45]

5:
1/2 , 3/5 , .606 , 13/20 , 66%
6:
0.09 , 1/10 , 12% , .13 , 3/20

8 0

3 years ago

50 Pts!! Brainliest!! Answer ASAP, thx.

Y_Kistochka [10]

Answer:

1/625

Step-by-step explanation:

When you multiply, you add the exponents. You now have $6^{-4}$ . This is equivalent to $\frac{1}{5^{4} }$ which equals 1/625.

6 0

3 years ago

Read 2 more answers

Whats 1+1 because i think its 11

NNADVOKAT [17]

Answer:

Step-by-step explanation:

when i have one fruit and you have another fruit and we add them together they are gonna b two

8 0

2 years ago

PLEASE HELP!! Algebra 2 Review !!

Kamila [148]

Answer:

$\frac{28-3i}{26}$

Step-by-step explanation:

For #2, remember that $i=\sqrt{-1}$ , so $i^{2}=-1$ Also, (a+b)(a-b), where a and b are any numbers, (a+b)(a-b)= $a^2-b^2$ . Now, to simplify, or radicalize, a number with surds in the denominator, you have to multiply the denominator by its conjugate. If there is a complex number $a+bi$ , where a and b are any numbers, the conjugate is always $a-bi$ . Lets apply these rules. The conjugate of 4+6i is 4-6i, so do this: