Answer:
an exponential function
Step-by-step explanation:
An exponential function has the characteristic that it increases by a constant factor when the independent variable increases by a constant increment. The function you describe is an exponential function.
__
It can be written in the form ...
f(t) = f(0)·2^(t/d) . . . . . where d is the doubling time
Using it's concepts, it is found that for the function
:
- The vertical asymptote of the function is x = 25.
- The horizontal asymptote is y = 5. Hence the end behavior is that
when
.
<h3>What are the asymptotes of a function f(x)?</h3>
- The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
- The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity. They also give the end behavior of a function.
In this problem, the function is:

For the vertical asymptote, it is given by:
x - 25 = 0 -> x = 25.
The horizontal asymptote is given by:

More can be learned about asymptotes at brainly.com/question/16948935
#SPJ1
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
5s-100-2s=4s-20-2s
3s-100. = 2s-20
+20. +20
3s-80. = 2s
-3s. -3s
-80. = -1s
-1 -1
80. = s
s=80