let's firstly convert the decimal to a fraction, namely a rational.
so we have one decimal, thus let's use 1 zero at the denominator and lose the dot on the numerator.
so then, what rationals are between -8/5 and +8/5? well, a huge amount hmmm let's pick two.
![\bf \boxed{-\cfrac{8}{5}}\rule[0.35em]{9.5em}{0.25pt}~~-\cfrac{3}{5}~~\rule[0.35em]{2em}{0.25pt}0\rule[0.35em]{10em}{0.25pt}~~\cfrac{6}{5}~~\rule[0.35em]{2.5em}{0.25pt}\boxed{\cfrac{8}{5}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B-%5Ccfrac%7B8%7D%7B5%7D%7D%5Crule%5B0.35em%5D%7B9.5em%7D%7B0.25pt%7D~~-%5Ccfrac%7B3%7D%7B5%7D~~%5Crule%5B0.35em%5D%7B2em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D~~%5Ccfrac%7B6%7D%7B5%7D~~%5Crule%5B0.35em%5D%7B2.5em%7D%7B0.25pt%7D%5Cboxed%7B%5Ccfrac%7B8%7D%7B5%7D%7D)
At lease means ≥sign to be used on left of given amount
Here it's 49
So it's
Now
If left paper be x
The inequality is
Option C
Answer:
C: 31 m/s
Step-by-step explanation:
It falls from a dam.
Thus;
Initial velocity; u = 0 m/s
Height; h = 49 m
Acceleration due to gravity is 9.8 m/s²
To solve for the final velocity this, we will use Newton's third equation of motion;
v² = u² + 2gh
v² = 0² + 2(9.8 × 49)
v = √960.5
v ≈ 31 m/s
First if a>0 sqrta=a^(1/2), and (a^p)^q=a^pxq
sqrt(a^4/9)= (a^4/9)^1/2=a^4/9x1/2=<span>a^(2/9)
so the answer is </span><span>a^(2/9)</span>
Answer:
Just copy and paste the question and there will be at least 2 to 3 sites with the answer
Step-by-step explanation:
