Answer:
0 ≤ t ≤ 5.
Step-by-step explanation:
In the function 
, 
is the independent variable. The domain of 
is the set of all values of that this function can accept.
In this case, is defined in a real-life context. Hence, consider the real-life constraints on the two variables. Both time and volume should be non-negative. In other words,
.
.
The first condition is an inequality about , which is indeed the independent variable.
However, the second condition is about , the dependent variable of this function. It has to be rewritten as a condition about .
![\begin{aligned} f(t) &\ge 0 &&\text{Assumption} \cr 80000 - 16000\, t& \ge 0 && \text{Definition of} ~ f \cr 80000 & \ge 16000\, t && \begin{aligned}&\text{Add $16000\, t$} \\[-0.5em] & \text{to both sides of the inequality}\end{aligned} \cr 5 &\ge t &&\begin{aligned}&\text{Divide both sides of} \\[-0.5em] & \text{the inequality by $16000$}\end{aligned} \cr t &\le 5 && \text{Flip the inequality}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20f%28t%29%20%26%5Cge%200%20%26%26%5Ctext%7BAssumption%7D%20%5Ccr%2080000%20-%2016000%5C%2C%20t%26%20%5Cge%200%20%26%26%20%5Ctext%7BDefinition%20of%7D%20~%20f%20%5Ccr%2080000%20%26%20%5Cge%2016000%5C%2C%20t%20%26%26%20%5Cbegin%7Baligned%7D%26%5Ctext%7BAdd%20%2416000%5C%2C%20t%24%7D%20%5C%5C%5B-0.5em%5D%20%26%20%5Ctext%7Bto%20both%20sides%20of%20the%20inequality%7D%5Cend%7Baligned%7D%20%5Ccr%205%20%26%5Cge%20t%20%26%26%5Cbegin%7Baligned%7D%26%5Ctext%7BDivide%20both%20sides%20of%7D%20%5C%5C%5B-0.5em%5D%20%26%20%5Ctext%7Bthe%20inequality%20by%20%2416000%24%7D%5Cend%7Baligned%7D%20%5Ccr%20t%20%26%5Cle%205%20%26%26%20%5Ctext%7BFlip%20the%20inequality%7D%5Cend%7Baligned%7D)
.
Hence, t ≤ 5.
Combine the two inequalities to obtain the domain:
0 ≤ t ≤ 5.
Would it be 4(1)^2+2(2(1)+2)^2=100
Let x be larger integer so x-3 is the smaller
so x^2+ x-3 = 87
x^2 +x -90=0 and factor
(x-9)(x+10)=0
x=9 or x=-10 but x must be positive
so x=9 and smaller number is 6
Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
