The restrictions on the variable of the given rational fraction is y ≠ 0.
<h3>The types of numbers.</h3>
In Mathematics, there are six (6) common types of numbers and these include the following:
- <u>Natural (counting) numbers:</u> these include 1, 2, 3, 4, 5, 6, .....114, ....560.
- <u>Whole numbers:</u> these comprises all natural numbers and 0.
- <u>Integers:</u> these are whole numbers that may either be positive, negative, or zero such as ....-560, ...... -114, ..... -4, -3, -2, -1, 0, 1, 2, 3, 4, .....114, ....560.
- <u>Irrational numbers:</u> these comprises non-terminating or non-repeating decimals.
- <u>Real numbers:</u> these comprises both rational numbers and irrational numbers.
- <u>Rational numbers:</u> these comprises fractions, integers, and terminating (repeating) decimals such as ....-560, ...... -114, ..... -4, -3, -2, -1, -1/2, 0, 1, 1/2, 2, 3, 4, .....114, ....560.
This ultimately implies that, a rational fraction simply comprises a real number and it can be defined as a quotient which consist of two integers x and y.
<h3>What are
restrictions?</h3>
In Mathematics, restrictions can be defined as all the real numbers that are not part of the domain because they produces a value of 0 in the denominator of a rational fraction.
In order to determine the restrictions for this rational fraction, we would equate the denominator to 0 and then solve:
23/7y;
7y = 0
y = 0/7
y ≠ 0.
Read more on restrictions here: brainly.com/question/10957518
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Complete Question:
State any restrictions on the variables 23/7y
Area of square = L×W
L×W = 13×13
L×W = 169
Answer:
sinθ = 
Step-by-step explanation:
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Answer:
Hey you were the one I helped before right?
Well I guess I can't stop u from giving doubts. Anyway, I'll give I the answer for this one as well.
Hope this helped (again lol), U can ask for help anytime.
Don't forget to give thanks.
Part a.
Fixed charge for the month: $30
Charge per hour: $0.50 daytime
Charge per hour: $0.25 nights and weekends
Let's say in one month, a person parks for d hours of daytime and n hours of nights and weekends.
The total monthly charge would be
cost = 30 + 0.50d + 0.25n
Now let's see what Trent did.
He parked for 47 hours in one month.
h of the 47 hours are nights and weekends.
Let x = number of daytime hours.
x + h = 47
x = 47 - h
He parked h hours of night and weekends, and he parked 47 - h hours of daytime.
Now we use h for night and weekend hours and 47 - h for daytime hours in the expression above.
cost = 30 + 0.50d + 0.25n
cost = 30 + 0.50(47 - h) + 0.25h
Answer to part a.: 30 + 0.50(47 - h) + 0.25h
Part b.
We are told the actual number of night and weekend hours, which we called h above, is 12. h = 12.
Now we use the cost expression we found in part a. with 12 in for h.
cost = 30 + 0.50(47 - h) + 0.25h
cost = 30 + 0.50(47 - 12) + 0.25(12)
cost = 30 + 0.50(35) + 0.25(12)
cost = 30 + 17.50 + 3
cost = 50.5
Answer to part b.: $50.50
