Answer:
The smaller number is 127
Step-by-step explanation:
Lets write the given problem in equation form
sum of consecutive numbers n and n + 1 = n + n + 1 = 2n + 1
now we find twice of the sum of consecutive numbers n and n + 1
2*(2n + 1) = 4n + 2
given that
twice the sum of consecutive numbers n and n + 1 is 510
thus,
4n + 2 = 510
=> 4n = 510 -2 = 508
=> n = 508/4 = 127
Thus, the numbers are n = 127
n+1 = 127 + 1 = 128
the smaller number is 127
1. 81 1/2 = 163/2
2. 32 1/5 - 64 1/3 =
483/15 - 965/15 =
-482/15 or -32 2/15
3. 16 1/4 = 65/4
4. 49 1/2 + 27 1/3 =
99/2 + 82/3 =
297/6 + 164/6 =
461/6 or 76 5/6
Bet what is it?................
Range is the outputs possible
w(r(x))=(2-x^2)-2
w(r(x))=2-x^2-2
w(r(x))=-x^2
therefor the range is from 0 to -∞
