Answer:
i belive its 35/8 3.14
Step-by-step explanation:
sorry if im wrong
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
First we will find the x coordinate.
x1+x2 / 2 = the x coordinate of the midpoint.
-9 + x2 / 2 = 10
x2/2 -4.5 = 10
x2/2 = 14.5
x2 = 29
Now the y coordinate:
y1+y2 /2 = y coordinate of the midpoint
7 + y2 / 2 = -3
y2/2 + 3.5 = -3
y2/2 = -6.5
y2 = -13
So the other endpoint is (29,-13)
I hope i am right :)
Answer:
i got 1200 not sure ifs right tho
Step-by-step explanation:
Answer:
Your just going to keep adding 3 and the 2 is there because you had negative -1 and then you added 3 so hope this helps probably not what you looking for but hope this helps. Have a good day.
Step-by-step explanation:
