I wrote the domain and range in interval notation, not sure if that’s how you’re asked to do it. This is how I was taught to state the domain/range of a graph.
<span>Ending Amt = Bgng Amt * e ^-0.03t
In this equation, the "-0.03" is the decay factor or "k"
We can now solve for half-life by this equation:
</span>t = <span>(<span>ln [y(t) ÷ a]<span>)<span> ÷ -k (we can say beginning amount = 200 and ending amount = 100
</span></span></span></span>t = <span>(<span>ln [200 ÷ 100]<span>)<span> ÷ -k
</span></span></span></span>t = <span>(<span>ln [2]<span>)<span> ÷ -k
</span></span></span></span>t = 0.69314718056<span> ÷ --.03
t =</span><span><span><span> 23.1049060187
</span>
about 23 years
</span></span>
Isn’t it h?
extra words so i can actually post this reply
First, think of it as 3 different expressions.Go through them one at a time. There is:
-2a(a+b-5) ] +3(-5a+2b) ] +b(6a+b-8)
Multiply everything out.] Multiply everything out. ]Multiply everything out
-2a x a = -2a² ] +3 x -5a = -15a ] +b x 6a = 6ab
-2a x b = -2ab ] +3 x 2b = 6b ] +b x b = b²
-2a x -5 = 10a ] -15a + 2b = -15a+2b ] +b x -8 = -8b -2a² + -2ab + 10a = ] ] 6ab + b² + -8b =
-2a²+-2ab+10a ] ] <span>6ab+b²+-8b
</span>Now add everything up, but before that, remember these algebraic rules:
∞A.S.S. Add Same Signs, meaning positive add positive or negative add negative equals a positive number or 0. E.g 1+1=1 and -1+-2= 1
∞S.I.D. Subtract If Different, meaning a positive number add a negative number always equals a negative number or 0. E.g 1 + -2 = -1
∞You can only add up "similar terms", meaning you can add 'terms' ending in 'a' for example with another 'term' ending in 'a'. However. you cannot add a term ending in 'a' to an 'ab' or an 'a²'.
Following these rules, -2a²+-2ab+10a+-15a+2b+<span>6ab+b²+-8b =
</span>4ab-5a-6b+b²-2a²
Hope that helps you and that it wasn't too tricky to understand. :)
