Here we will use the trial and error method.
- We will try putting different values of x.
<h3 /><h3>1st of all</h3><h3>x=1</h3>





<h2>Now,</h2><h3>x=2</h3>




<h2>Again,</h2><h3>x=3</h3>




<h3>☣Hence, The value of X as 3 satisfies the equation!</h3>
-3n < 81...divide both sides by -3, change the inequality sign
n > -81/3
n > - 27 <==
** in an inequality, when dividing/multiplying by a negative number, the inequality sign changes
Answer:
the last one
Step-by-step explanation:
it's basically finding the area for a cube and then cutting it in half, so multiply it all together and then divide by 2. sorry if I'm wrong lol
Say 36% of 100 is 36. so knowing half of 36 is 18, take half of 100, 50.
so 18/50 = 36
The answer is 50.