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: The first one
Step-by-step explanation:
If you take a look at the light pink squares, you can count three on the top and three on the bottom. Therefore, when looking at it from the top, you’ll see the first one.
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Even a brainliest if you feel like it! ;)
Answer:
6, 5, 1, 7, 9, 2, 4, 8, 3.
Step-by-step explanation:
(from top to bottom, this should be right)
Answer:
15 (3+1)
Step-by-step explanation:
Hope this helps :)
Answer:
T F → F
Step-by-step explanation:
2+2 does equal 4 so its true.
but 7+1 does <em>not</em><em> </em>equal 7, it equals 8. So its fasle.
im not sure i did this this right so if i did it wrong im sorry love (-ω-;)