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:
lol
Step-by-step explanation:
(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.
Option B
<u>Step-by-step explanation:</u>
Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:
First look at where we have intersection points, In first quadrant & in third quadrant.
<u>At first quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)
<u>At Third quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y= -2.0. So, other point is (-1.2,-2.0).