You can see how this works by thinking through what's going on.
In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97
What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.
So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.
So the population after x years is 1200 * 0.97x
Answer:
Step-by-step explanation:
h(x)=44x+8-(30x+15)
=44x+8-30x-15
=44x-30x+8-15
=14x-7
B= the base of the object
<u>Answer:</u>
a) 3.675 m
b) 3.67m
<u>Explanation:</u>
We are given acceleration due to gravity on earth =
And on planet given = 
A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula </u>
Where H = max jump height,
v0 = velocity of jump,
Ø = angle of jump and
g = acceleration due to gravity
Considering velocity and angle in both cases
Where H1 = jump height on given planet,
H2 = jump height on earth = 0.75m (given)
g1 = 2.0
and
g2 = 9.8
Substituting these values we get H1 = 3.675m which is the required answer
B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by </u>
which is due to projectile motion of ball
Now h = max height,
v0 = initial velocity = 0,
t = time of motion,
a = acceleration = g = acceleration due to gravity
Considering t = same on both places we can write
where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations
substituting h2 = 18m, g1 = 2.0 and g2 = 9.8
We get h1 = 3.67m which is the required height
