Let x and y be your two consecutive whole numbers
x < sqrt(142) < y
x^2 < 142 < y
So, we are looking for x and y such that x^2 < 142 and y^2 > 142.
The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.
11 and 12 are consecutive.
11^2 = 121 < 142 < 144 = 12^2
Thus, 11 and 12 are your numbers
Answer:
b
Step-by-step explanation:
Answer: Do the math, the positive solution
t = 5.0495 seconds (If it's wrong sorry, I havent done a problem like this one in awhile!)
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let 
y=2x-10
Switch x and y,
x=2y-10
x+10=2y
(x/2)+5=y
So 
Answer:
C and D
Step-by-step explanation:
You can tell if equations have no solution if the variables on both sides of the equation are the same.
Choice C is 10+6x=15+9x-3x. You would combine 9x and -3x, turning the equation into 10+6x=15+6x. Since there is a 6x on both sides, Choice C would have no solution.
The same for Choice D. When you simplify everything there is a 3x on both sides, so it would have no solution.