Altho' I can easily guess what you're supposed to do here, I must point out that you haven't included the instructions for this problem.
I'll help you by example. Let's look at the first problem:
"Evaluate 6(z-1) at z-4."
Due to "order of operations" rules, we must do the work inside the parentheses FIRST. Replace the z inside (z-1) with "-4". We obtain
6(-4-1) = 6(-5) = -30 (answer.)
Your turn. Try the next one. If it's unclear, as questions.
Answer:
Length = 42.5 cm
Step-by-step explanation:
length of rectangle = 5 + 3w
So l = 5 + 3w where w = 12.5
Equation:
L = 5 + 3(12.5)
Solving the Equation:
L = 5 + 3(12.5)
L = 5 + 36 + 1.5
L = 42.5
Length: 42.5 cm
If my answer is incorrect, pls correct me!
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-Chetan K

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 

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