prove that A+B+B+B+80+90=100
hey its wrong!!!!
it is a logical question
not a mathematical question
Answers
Answered by
3
A+B+B+B+80+90=100
A+3B=100-90-80
A+3B=-70
#let A be -10–1
#let B be -20–2
#from 1 and 2 we get
(-10)+3(-20)=-70
-10-60=-70
#-70=-70
#lhs = rhs
therefore A+B+B+B+80+90=100
A+3B=100-90-80
A+3B=-70
#let A be -10–1
#let B be -20–2
#from 1 and 2 we get
(-10)+3(-20)=-70
-10-60=-70
#-70=-70
#lhs = rhs
therefore A+B+B+B+80+90=100
wronggggg!!!!!!!!!!!!
Answered by
0
Answer:
ALL ARE WRONG!!!!!!!!!!!!!!!!
Step-by-step explanation:
PROVED AS FOLLOWS:
Akad Bakad Bambe Bo 80 90 pure 100
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log [1/(10^10)] = -10 log(10) = -10 ->A
log [1/(10^20)] = -20 log(10) = -20 ->B
Now,
So, LHS: A + 3B + 80 + 90
=> -10 + 3(-20) + 80 + 90
=> 100 -> RHS
Hence, A is log[1/(10^10)] and B is log[1/(10^20)] , no negative numbers though!
Thanks for asking:)