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chubhunter [2.5K]
2 years ago
10

How do I solve “1+2+3+4+5+…100=“ A. 1010 B. 5050 C. 5000 D. 1000

Mathematics
1 answer:
irina [24]2 years ago
5 0

Answer:

5050

Step-by-step explanation:

we know that 100/2 is 50 and 50 x 100 is 5000

so now another 50 is remaining but we can't multiply but add so

1+2+3+4+5...100 = in simple form= 50x100+50=5050//

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I really need help with this problem steps
melamori03 [73]
Solve for y by setting one equation =to X after that you can substitute the equation into the other one let's set the first one in X= FORM
X-7y=10
-2x+14y=-20

Add 7y to both sides of the first equation to let x stand alone
X=7y+10
Now you can substitute x on the second equation
-2 (7y+10)+14y=-20
Distribute
-14y-20+14y=-20
Add and simplify
Us cancel out =0
-20=-20
0=0
They are the same equations you can also divide the second equation by -2 which would make it look like this
X-7y=10
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Let me know if I have answered your question
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