Answer:
1.994
Step-by-step explanation:
"Percent" means "per 100" or "over 100". To convert 199.4% to a decimal rewrite 199.4 percent in terms of per 100 or over 100.
199.4% = 199.4 over 100 or,
199.4%=199.4100
199.4 over 100 is the same as 199.4 divided by 100. Completing the division we get:
199.4 ÷ 100 = 1.994
Therefore, we have shown that
199.4% = 1.994
Simplified Conversion:
Remove the percent sign % and divide by 100.
199.4 ÷ 100 = 1.994
Shortcut Conversion:
Move the decimal point 2 places to the left and remove the percent sign %
199.4% becomes 1.994
When you say comparing do you mean dividing or just comparing
Answer:
C. 1 inch long, 1 inch wide, 1 inch high
Step-by-step explanation:
In this instance, unit just means a singularity in measurement
Answer:
<em>LCM</em> = ![8y^{4}(y+ 10)^{2}(y + 8)](https://tex.z-dn.net/?f=8y%5E%7B4%7D%28y%2B%2010%29%5E%7B2%7D%28y%20%2B%208%29)
Step-by-step explanation:
Making factors of ![8y^{6}+ 144y^{5}+ 640y^{4}](https://tex.z-dn.net/?f=8y%5E%7B6%7D%2B%20144y%5E%7B5%7D%2B%20640y%5E%7B4%7D)
Taking
common:
![\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)](https://tex.z-dn.net/?f=%5CRightarrow%208y%5E%7B4%7D%20%28y%5E%7B2%7D%2B%2018y%2B%2080%29)
Using <em>factorization</em> method:
![\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)](https://tex.z-dn.net/?f=%5CRightarrow%208y%5E%7B4%7D%20%28y%5E%7B2%7D%2B%2010y%20%2B%208y%20%2B%2080%29%5C%5C%5CRightarrow%208y%5E%7B4%7D%20%28y%20%28y%2B%2010%29%20%2B%208%28y%20%2B%2010%29%29%5C%5C%5CRightarrow%208y%5E%7B4%7D%20%28y%2B%2010%29%28y%20%2B%208%29%29%5C%5C%5CRightarrow%20%5Cunderline%7B2y%5E%7B2%7D%7D%20%5Ctimes%20%204y%5E%7B2%7D%20%5Cunderline%7B%28y%2B%2010%29%7D%28y%20%2B%208%29%29%20.....%20%281%29)
Now, Making factors of ![2y^{4} + 40y^{3} + 200y^{2}](https://tex.z-dn.net/?f=2y%5E%7B4%7D%20%2B%2040y%5E%7B3%7D%20%2B%20200y%5E%7B2%7D)
Taking
common:
![\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)](https://tex.z-dn.net/?f=%5CRightarrow%202y%5E%7B2%7D%20%28y%5E%7B2%7D%2B%2020y%2B%20100%29)
Using <em>factorization</em> method:
![\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)](https://tex.z-dn.net/?f=%5CRightarrow%202y%5E%7B2%7D%20%28y%5E%7B2%7D%2B%2010y%2B%2010y%2B%20100%29%5C%5C%5CRightarrow%202y%5E%7B2%7D%20%28y%20%28y%2B%2010%29%20%2B%2010%28y%20%2B%2010%29%29%5C%5C%5CRightarrow%20%5Cunderline%20%7B2y%5E%7B2%7D%20%28y%2B%2010%29%7D%28y%20%2B%2010%29%20%20%20%20%20%20%20%20............%20%282%29)
The underlined parts show the Highest Common Factor(HCF).
i.e. <em>HCF</em> is
.
We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:
![HCF \times LCM = p \times q](https://tex.z-dn.net/?f=HCF%20%5Ctimes%20LCM%20%3D%20p%20%5Ctimes%20q)
Using equations <em>(1)</em> and <em>(2)</em>:
![\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)](https://tex.z-dn.net/?f=%5CRightarrow%202y%5E%7B2%7D%20%28y%2B%2010%29%20%5Ctimes%20LCM%20%3D%202y%5E%7B2%7D%20%5Ctimes%20%204y%5E%7B2%7D%28y%2B%2010%29%28y%20%2B%208%29%20%5Ctimes%202y%5E%7B2%7D%20%28y%2B%2010%29%28y%20%2B%2010%29%5C%5C%5CRightarrow%20LCM%20%3D%202y%5E%7B2%7D%20%5Ctimes%20%204y%5E%7B2%7D%28y%2B%2010%29%28y%20%2B%208%29%20%5Ctimes%20%28y%20%2B%2010%29%5C%5C%5CRightarrow%20LCM%20%3D%208y%5E%7B4%7D%28y%2B%2010%29%5E%7B2%7D%28y%20%2B%208%29)
Hence, <em>LCM</em> = ![8y^{4}(y+ 10)^{2}(y + 8)](https://tex.z-dn.net/?f=8y%5E%7B4%7D%28y%2B%2010%29%5E%7B2%7D%28y%20%2B%208%29)
Answer: Together the pool can be drained in 2 hours and 18 minutes.