Answer:
let the numbers be, x and x+1 ( as it is consecutive number the next number will have 1 more than the number )
sum = 41
so, x + ( x + 1 ) = 41
x + x + 1 = 41
2x + 1 = 41
2x = 41 -1 = 40
2x = 40
x = 40/2 = 20
x = 20
x + 1 = 20 + 1 = 21
so the consecutive numbers whose sum is 40 are, 20 and 21
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0.015 or 3/200
15 divided by 1000 is 0.015
15/5 is 3 1000/5 is 200
You would plug in 4 for x.
f(4)= (4) - 6 = -2
