Post the problem from the paper
When the height is 0 that is when it hits the ground
solve
0=-16t²+25x+3
cant factor so use quadratic formula
for
0=ax²+bx+c
so for
0=-16t²+25t+3
a=-16
b=25
c=3
so
it has to be a positive time
so it will hit the ground after
seconds
1] Given that the value of x has been modeled by f(x)=12500(0.87)^x, then:
the rate of change between years 1 and 5 will be:
rate of change is given by:
[f(b)-f(a)]/(b-a)
thus:
f(1)=12500(0.87)^1=10875
f(5)=12500(0.87)^5=6230.3
rate of change will be:
(6230.3-10875)/(5-1)
=-1161.2
rate of change in years 11 to 15 will be:
f(11)=12500(0.87)^11=2701.6
f(15)=12500(0.87)^15=1,547.74
thus the rate of change will be:
(1547.74-2701.6)/(15-11)
=-288
dividing the two rates of change we get:
-288/-1161.2
-=1/4
comparing the two rate of change we conclude that:
The average rate of change between years 11 and 15 is about 1/4 the rate between years 1 and 5.
The answer is D]
2] Given that the population of beavers decreases exponentially at the rate of 7.5% per year, the monthly rate will be:
monthly rate=(n/12)
where n is the number of months
=7.5/12
=0.625
This is approximately equal to 0.65%. The correct answer is A. 0.65%
Answer:
B. -y = -x
Step-by-step explanation:
When the values on both sides of the equal sign are changed, it becomes a negative number.
Answer:
The amount of invested money after a period of 4 years:
A = 200 x (1 + 9/100)^4 = 282.32 dollar
Hope this helps!
:)
