The inventory account expected to have by December 31 is more than $5800. Option C
<h3>How to calculate the end inventory</h3>
The formula for end inventory is given as ;
Ending inventory = Beginning inventory + net purchases –sales
Beginning inventory = $5800
Net purchases = $65000
Sales = $112000
Put into the formula
Ending inventory = $ 
Add first,
Ending inventory = $ 
Ending inventory = $ -41, 200
Thus, the inventory account expected to have by December 31 is more than $5800. Option C
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I like to reduce everything by half if I don't already know the answer.
It just so happens that is the reduced fraction.
All I did was halve 18, which is 9, and halve 34, which is 17.
Answer: 
Step-by-step explanation:
Given
Max is driving at a speed of 
It took him three-quarters of a second i.e. 
Speed in meter per second is 
Distance is given by
Reaction distance is
Answer:
h(8q²-2q) = 56q² -10q
k(2q²+3q) = 16q² +31q
Step-by-step explanation:
1. Replace x in the function definition with the function's argument, then simplify.
h(x) = 7x +4q
h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q
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2. Same as the first problem.
k(x) = 8x +7q
k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q
_____
Comment on the problem
In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...
h(8q² -2q) = 56q² -14q +4z
and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.
In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.
Answer:
6
Step-by-step explanation:
fog(x) = √3(x-6)
Domain => 3(x-6) ≥ 0
<=> x-6≥ 0
=> x ≥ 6
so, the smallest number is 6
