Answer:
A
Step-by-step explanation:
This is the answer because the others don't make sense to me at all...
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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.
Answer:
x^2+8x+16
(x+4)^2
Step-by-step explanation:
x^2 +8x
Take the coefficient of the x term
8
Divide by 2
8/2 =4
Square it
4^2 =16
x^2+8x+16
We can factor it into
(x+8/2) ^2
(x+4)^2
Answer:
See below.
Step-by-step explanation:
This is how you prove it.
<B and <F are given as congruent.
This is 1 pair of congruent angles for triangles ABC and GFE.
<DEC and <DCE are given as congruent.
Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.
This is 1 pair of congruent angles for triangles ABC and GFE.
Now you need another side to do either AAS or ASA.
Look at triangle DCE. Using the fact that angles DEC and DCE are congruent, opposite sides are congruent, so segments DC and DE are congruent. You are told segments DF and BD are congruent. Using segment addition postulate and substitution, show that segments CB and EF are congruent.
Now you have 1 pair of included sides congruent ABC and GFE.
Now using ASA, you prove triangles ABC and GFE congruent.
