The answer is -1....................
Step-by-step explanation: To solve this absolute value inequality,
our goal is to get the absolute value by itself on one side of the inequality.
So start by adding 2 to both sides and we have 4|x + 5| ≤ 12.
Now divide both sides by 3 and we have |x + 5| ≤ 3.
Now the the absolute value is isolated, we can split this up.
The first inequality will look exactly like the one
we have right now except for the absolute value.
For the second one, we flip the sign and change the 3 to a negative.
So we have x + 5 ≤ 3 or x + 5 ≥ -3.
Solving each inequality from here, we have x ≤ -2 or x ≥ -8.
Answer:
A. 25 < (x – 1)² + y² and 16 > x² + (y + 4)²
Step-by-step explanation:
the solutions are in the outside of the bigger circle, but inside of the smaller circle
Y= -1 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
well, let's first notice, all our dimensions or measures must be using the same unit, so could convert the height to liters or the liters to centimeters, well hmm let's convert the volume of 1000 litres to cubic centimeters, keeping in mind that there are 1000 cm³ in 1 litre.
well, 1000 * 1000 = 1,000,000 cm³, so that's 1000 litres.
![\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=1000000~cm^3\\ h=224~cm \end{cases}\implies \stackrel{cm^3}{1000000}=\pi r^2(\stackrel{cm}{224}) \\\\\\ \cfrac{1000000}{224\pi }=r^2\implies \sqrt{\cfrac{1000000}{224\pi }}=r\implies \cfrac{1000}{\sqrt{224\pi }}=r\implies \stackrel{cm}{37.7}\approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%20V%3D%5Cpi%20r%5E2%20h~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D1000000~cm%5E3%5C%5C%20h%3D224~cm%20%5Cend%7Bcases%7D%5Cimplies%20%5Cstackrel%7Bcm%5E3%7D%7B1000000%7D%3D%5Cpi%20r%5E2%28%5Cstackrel%7Bcm%7D%7B224%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1000000%7D%7B224%5Cpi%20%7D%3Dr%5E2%5Cimplies%20%5Csqrt%7B%5Ccfrac%7B1000000%7D%7B224%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Ccfrac%7B1000%7D%7B%5Csqrt%7B224%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B37.7%7D%5Capprox%20r)
now, we could have included the "cm³ and cm" units for the volume as well as the height in the calculations, and their simplication will have been just the "cm" anyway.
