Answer:
Step-by-step explanation:
<u>Firstly we'll factorize the terms:</u>
45x³ = 3 * 3 * 5 * x * x * x
24x² = 2 * 2 * 2 * 3 * x * x
16x = 2 * 2 * 2 * 2 * x
<u>The factor which is common in all three terms is:</u>
= x
Hence GCF = x
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Let's make things easier by simplifying things.
y = 8 and x = 3 is more likely to be understood as a ratio. So for the rest of the answer, their relationship would be represented as y:x
Thus: y:x = 8:3
The problem would be finding y when x = 45
Let us proceed on using the previous equation and substitute x with 45 which would look like this:
y:45 = 8:3
Ratios can also be expressed as fractions which would make things more understandable and easy to solve. So the new form of our equation would be like this:
y/45 = 8/3
Then we proceed with a cross multiplication where the equation becomes like as what is shown below:
3y = 45 * 8
From there, you can solve it by multiplying 45 and 8 then dividing the product with 3 to get y
3y = 360
y = 120
Another way of looking at the problem, especially problems like these, is to take the whole question or statement as an equation. it would probably look like this:
y = 8 when x = 3 : y = ? when x = 45
This would make you understand what approach you can use to solve the given problem.
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
brainly.com/question/4318844
#SPJ4
Three numbers between 3.65 and 3.66 could be:
1) 3.651
2) 3.655
3) 3.657
4) 3.659
Hope this helps! :)
