tli8322 tli8322
  • 04-08-2022
  • Mathematics
contestada

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

Respuesta :

jetsungyembo
jetsungyembo jetsungyembo
  • 04-08-2022

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//

Answer Link

Otras preguntas

Marty ate 4/6 of his pizza and luis ate 5/6 of his pizza. Marty ate more of his pizza. How is this possible
from 3 o'clock to 3:45 p.m. the min tuns how many degrees
Because of his original approach and forms, Wilder may be called a(n) ______.
Two fractions have the sane denominator. Which is the greater fractionthe fraction with the greater numerator oe the less numerator
The amount of kinetic energy in an objcet deoends on 2things its ? plz tell me
define probability. give an example of the probability of a simple event.
A mutation that causes changes in the amino acid sequence downstream of the mutation is called: point mutation, translocation, substitution, or frameshift mutat
Which of the following is contained in the header of a memo? From Where Why Who
Inflections change the grammatical meaning of ______. prefixes suffixes root words
Similarity between heat and temperature