<h3>
Answer: D) 144 square cm</h3>
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Explanation:
A square has all four sides that are the same length. So all four sides are 12 cm
area = side*side = 12*12 = 144 cm^2
Or you could say
area = (side)^2 = (12)^2 = 144 cm^2
Squaring a number is the same as multiplying it by itself. Hence the name "square" refers to a side length pairing up with its corresponding area value.
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Side note: The perimeter is 4*(side) = 4*12 = 48 cm. It's possible that choice B is put in there as a trick answer.
A) √50 = 2√25 . Her mistake was that she confused the radicand (25) & the outside factor to the radicand that is 2. The write answer is 5√2 & not the opposite
b) Find 2 numbers where their square nears the lower & the upper end of 50.
For instance 7² = 49 & 8² =64 are the nearest lower & upper numbers that encompass 50
Hence we can write the following inequality:
49<50<64 & √49<√50√64 ==> 7<√50<8
So the √50 is nearer to 7² rather than 8².
Then let's try 7.1² =50.41 Which is an acceptable approximation
let's first off notice that, on the 2), the sector is really half of the whole circle, and on 3) the sector is one quarter of the whole circle.
now, on 2) AB is the diameter of 4 units, therefore it has a radius of 2, or half that.
![\bf \boxed{2} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies A=\pi 2^2\implies A=4\pi \\\\\\ \stackrel{\textit{half of that}}{A=2\pi}\implies A=\stackrel{\textit{rounded up}}{A=6.3~ft^2} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%202%5E2%5Cimplies%20A%3D4%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%7D%7D%7BA%3D2%5Cpi%7D%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D6.3~ft%5E2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \boxed{3} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=20 \end{cases}\implies A=\pi 20^2\implies A=400\pi \\\\\\ \stackrel{\textit{one quarter of that}}{A=100\pi }\implies \stackrel{\textit{rounded up}}{A=314.2~in^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D20%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2020%5E2%5Cimplies%20A%3D400%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7BA%3D100%5Cpi%20%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D314.2~in%5E2%7D)
Answer:
Simple!
Step-by-step explanation:
Multiply 5 into each term into the parenthesis.
5-`10g + 20h
Answer:
hey i did this on my quiz trust me
Step-by-step explanation
12,362.70 trust me i did this i just dont know exactly how to explain this
