The numbers are 97, and 99.
Let the number be x.15 * x= x+196Then, 15x - x =196‹=› 14 x= 196‹=› x = 14.
132 = 169 < 184 < 196 = 142 So 13 < √184 < 14 So estimating from consecutive whole numbers would give the answer "13 and a bit". But you can do better than that by interpolating. The square-root function is, obviously, not linear but you will not go too far wrong between consecutive integers (unless you are dealing with very small number). 184 is (184-169)/(196-169) = 15/27 of the distance between 132 and 142 so the square root is approx 15/27 = 5/9 of the way from 13 to 14 ie 13+5/9 = 13.555... recurring. In fact, the correct value is approx 13.5647 so the estimate is accurate to 2 decimal places.
Let the number be 'm' & 'm + 2' Hence their squares are m^2 & (m+2)^2 There sum is m^2 + (m+2)^2 = 340 Multiply out he brackets m^2 + m^2 + 4m + 4 = 340 . 2m^2 + 4m + 4 = 340 Divide both sides by '2' m^2 + 2m + 2 = 170 m^2 + 2m - 168 = 0 Factor (m - 12)(m + 14) = 0 Hence m = 12 & m+ 2 = 14 The two consecutive even numbers. 12^2 = 144 14^2 = 196 144 + 196 = 340 As required.
40% of 196= 40% * 196= 0.4 * 196= 78.4
97 + 99 = 196
They are: 97+99 = 196
The integers are 12 and 14 (144+196=340)
The numbers are 97, and 99.
10+(X)+(X+1)=196 10+2X+1=196 11+2X=196 2X=185 X=92.5 Therefore, ten plus the sum of two consecutive integers can not be even.
144 + 196 = 340 so integers are 12 & 14.
x^2 + (x+1 )^2 =365 x^2 +x^2 + 2x +1 =365 2x^2 + 2x = 364 x^2 + x = 182 = (x- 13)(x + 14) x =13, so integers are 13 and 14, squares 169 and 196, which total 365.
Integers are rational.
All integers are rational numbers.
13 & 14 (169 & 196)
Any one of infinitely many integers of the form 196*k where k is an integer.
The numbers are 195, 196, 197 and 198.