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:
It looks like c
Step-by-step explanation:
I'm not completely sure bit i did this last year
Umm this should be in physics but F is proportional to 1/r^2 I am assuming you want to just take the derivative in respect to r of both sides of the equation
F = (2.99x10^16)/r^2
So you do that and you get
dF/dr = -2(2.99 * 10 ^16)/ (r^3)
is this for calc class?