Answer:
Height of the cereal box = 18 in.
Step-by-step explanation:
A cereal box is in the shape of a cuboid.
And volume of a cuboid is represented by the expression,
Volume = Length × Width × Height
It's given that volume of the box = 432 in³
Length of the base = 2 in.
Width of the base = 12 in
By substituting these values in the expression,
432 = 2 × 12 × h
h = 
h = 18 in.
Therefore, height of the cereal box is 18 in.
Answer: 4logw (x^{2} - 6) - 1/3logw (x^{2} + 8)
Step-by-step explanation:
∠ M ≅ ∠ R: true
<span>VL ≅ LT: true
</span><span>Δ MLV can be rotated about point L to map it to Δ RLT. : false
</span><span>A series of rigid transformations of Δ MLV maps it to Δ RLT. : true </span>
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.