Answer:
3 ,sqrt(12), 4, sqrt(18), sqrt(23), 5, sqrt(27) sqrt(32), 6, 7, sqrt(50)
1. x=17
2. x=5
Step-by-step explanation:
When we are taking the square root of the numbers, the y would be in the same order as if we were not taking the square root
sqrt(12), sqrt(18), sqrt(23), sqrt(27) sqrt(32), sqrt(50)
Now we know that sqrt(9) =3, sqrt(16) =4, sqrt(25) =5 sqrt(36) = 6, sqrt(49( = 7
because these are perfect squares
Sliding these inbetween the numbers above
sqrt(9) ,sqrt(12), sqrt(16) sqrt(18), sqrt(23), sqrt(25) sqrt(27) sqrt(32), sqrt(36) sqrt(49) sqrt(50)
Replacing them with the whole numbers
3 ,sqrt(12), 4, sqrt(18), sqrt(23), 5, sqrt(27) sqrt(32), 6, 7, sqrt(50)
1. x^2 = 289
Take the square root of each side
sqrt(289) = sqrt(x^2)
±17 =x
We want x to be positive
17 =x
2. 125 = x^3
What number times itself * itself = 125
125 = x*x*x
x=5
5*5*5 = 125
