Since y=-1 when x=0 and we can write the equation (we'll turn it into an inequality later) as y=mx-1 from y=mx+b by plugging (0,1) in. Next, y equals 2 when x=1, so we plug those in to get 2=m*1-1. Adding 1 to both sides, we get 3=m, making our equation y=3x-1 (since y and x stay variables). Lastly, we turn it into an inequality, It seems to be inclusive to the line, so it's either
3x-1≤y or 3x-1≥y. Finding a random point in the inequality (4, 1), we plug it in to get 12-1=11, which is clearly larger than 1, so we get 3x-1≥y.
The answer is the first option: 1) > the quantity times minus all over
The explanation is shown below:
1. To solve this problem you must pply the following proccedure:
2. Move the term to the left member. As the variable is negative, multiply the expression by and change the direction of the sign:
I assume you meant N=100e^(t/10) so
200=100e^(t/10)
2=e^(t/10) taking the natural log of both sides
ln2=t/10
t=10ln2
t≈6.93 hours
check...
N=100e^(.693)≈200