<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days
Step-by-step explanation:
The given parameters are;
The half life of the radioactive substance = 45 days
The mass of the substance initially present = 6.2 grams
The expression for evaluating the half life is given as follows;

Where;
N(t) = The amount of the substance left after a given time period = 1 gram
N₀ = The initial amount of the radioactive substance = 6.2 grams
= The half life of the radioactive substance = 45 days
Substituting the values gives;




The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days
Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.
Answer:
a=4+2/x <-- 2 over x its a fraction
Step-by-step explanation:
X= a+b+c+bx/a(x-b) but I’m not completely sure