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lys-0071 [83]
3 years ago
7

Solve for X

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

Answer:

x = 1035 = MXXXV

Step-by-step explanation:

<u>Roman Numbers </u>

It's an ancient number system developed in Rome where numbers are represented by the letters {I,V,X,L,C,D,M}. The value of each are, respectively {1,5,10,50,100,500,1000}. If some letter is repeated, then it's value must be multiplied by the number of times the letter is present. For example, XXVII has two X's and two I's, so the conversion is 10+10+5+1+1=27.

If a less-valued letter appears before a greater one, it subtracts. For example CDI=-100+500+1=401

Our numbers are

MMMMMMMMCDXV=8*1000-100+500+10+5=8415

MDCCCLXIII = 1000+500+300+50+10+3=1863

DLXI = 500+50+10+1=561

XXVII = 20+5+2=27

We now operate in our numerical system

\displaystyle x = \frac{MMMMMMMMCDXV * MDCCCLXIII}{DLXI * XXVII}

\displaystyle x = \frac{8415 * 1863}{561 * 27}

\displaystyle x = \frac{15677145}{15147}

\displaystyle x = 1035

Converting back to roman

\boxed{\displaystyle x = MXXXV}

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Answer:

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Step-by-step explanation:

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Situation:
Rzqust [24]
N=NOe^-kt
N=mass at time t
NO = initial mass
k= 0.1476
t= time, in days

We are asked to find the half-life, which means you want to find how long it will take for half of the substance to decay/disappear (depending on situation). 

If we are looking for half-life, we can simply set N to half of NO (which we are given a value of 40grams for)
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N = 20
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plugging these values and the value given for k back into the equation you get:

20 = 40e^-0.1476(t)
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We can start by dividing 40 from both sides, and you get:
0.5 = e^-0.1476(t)

We have the exponential function "e".
To get rid of e, we can use natural log (ln)
if e^y=x then ln (x) = y
look back at our equation we can set
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Rewriting it in natural log form:
ln (0.5) = -0.1476(t)
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The last step is to simply divide both sides by -0.1476
therefore:
T = 4.696119
But it asks you for the answer to the nearest tenth (one place after decimal pt) so 
T (half life) = 4.7 days

Hope that helps :)
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