C) will not effected (the most frequent is still 600$)
$20×.25=$5
$20-$5=$15
$15×.15=$2.25
$15+$2.25=$17.25
$17.25 is your answer
Answer:
39
Step-by-step explanation:
Given the function :
h(t) = t² + 2
From t = 5 to t = 8
when, t = 5
h(5) = 5² + 2
h(5) = 25 + 2
h(t) at t = 5 ; equals 27
when, t = 8
h(5) = 8² + 2
h(5) = 64 + 2
h(t) at t = 8 ; equals 66
Net Change :
h(8). - h(5)
66 - 27 = 39

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>

</span><span>In notation we write respectively
</span>

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²

That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers 