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:
Y= 4x-3
Step-by-step explanation:
The y-intercept is -3,
the slope is 4:
rise over run= 4/1 or 4
so 4x-3
I hope this helps!
Answer: 78
Step by step: congruent side
You can, it's not as hard as you might think
simply multiply the speed by the fraction of time
23 × 1/5 = 23/5 = 4 3/5 = 4.6
Answer:
p+q = 38/15
Step-by-step explanation:
2/3+p = 1. p = 1 - 2/3
4/5 + q = 3. q = 3 - 4/5
p + q = 1 - 2/3 + 3 - 4/5
-2/3 - 4/5 (common denominator = 15)
4 - 10/15 - 12/15 = 4 - 22/15
4 - 22/15 = (4*15 - 22)/15
= 38/15
