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pychu [463]
2 years ago
6

Simplify and state the excluded value for (18r^2)/(12r)

Mathematics
1 answer:
Viefleur [7K]2 years ago
7 0

Answer:

216r^3

Step-by-step explanation:

18*r^2*12*r

216*r^3

216r^3

I hope it's help!

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If you double president Reagan’s age at the time of his first inauguration(69) and subtract his age at the time he died the resu
e-lub [12.9K]
It should be:

69(2)=138
138-48=90
5 0
2 years ago
PLS HELP ASAP. NO LINKS <br> When z= 9, the value of z2 + 1 is
elena55 [62]
Answer is 82

substitute z for 9

9^2 = 81+ 1= 82
6 0
2 years ago
When studying radioactive​ material, a nuclear engineer found that over 365​ days, 1,000,000 radioactive atoms decayed to 973 co
HACTEHA [7]

Answer:

A. number of decayed atoms = 73.197

Step-by-step explanation:

In order to find the answer we need to use the radioactive decay equation:

N(t)=N0*e^{kt} where:

N0=initial radioactive atoms

t=time

k=radioactive decay constant

In our case, when t=0 we have 1,000,000 atoms, so:

1,000,000=N0*e^{k*0}

1,000,000=N0

Now we need to find 'k'. Using the provied information that after 365 days we have 973,635 radioactive atoms, we have:

973,635=1,000,000*e^{k*365}

ln(973,635/1,000,000)/365=k

-0.0000732=k

A. atoms decayed in a day:

N(t)=1,000,000*e^{-0.0000732t}

N(1)=1,000,000*e^{-0.0000732*1}

N(1)= 999,926.803

Number of atoms decayed in a day = 1,000,000 - 999,926.803 = 73.197

B. Because 'k' represents the probability of decay, then the probability that on a given day 51 radioactive atoms decayed is k=0.0000732.

4 0
3 years ago
314,207 in standard form expanded form and word form
Gennadij [26K]
Three hundred fourteen thousand two hundread seven

Hope it helps!!
8 0
3 years ago
Read 2 more answers
Solve for x squared equals seventeen please <br>x^2 = 17​
Lena [83]

Answer: 4.123 (rounded) or sqrt(17)

Step-by-step explanation: Take the square root of 17.

7 0
3 years ago
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