Answer:
The year is 2020.
Step-by-step explanation:
Let the number of years passed since 2010 to reach population more than 7000000 be 'x'.
Given:
Initial population is, 
Growth rate is, 
Final population is, 
A population growth is an exponential growth and is modeled by the following function:

Taking log on both sides, we get:

Plug in all the given values and solve for 'x'.

So, for
, the population is over 700,000. Therefore, from the tenth year after 2010, the population will be over 700,000.
Therefore, the tenth year after 2010 is 2020.
Answer:
Step-by-step explanation:
x² + 4x - 1 - x =x² + 4x -x - 1 { bring the like terms together and add}
= x² + 3x - 1
<span>
the answer is 3.75 I'll explain ,</span><span>here: 1.5 times 2 equals 3 now go to a different section and divide 1.5 by 2 then add the two together it will look kind of like this:</span>
<span>(1.5*2)+(1.5/2)=3.75</span>