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
Answer:
0.548 = 5.48 * 
Step-by-step explanation:
in standard form you have to have most of the numbers after the decimal and only 1 behind it, in this case 5 is behind, and the rest infront.
Answer:
The answer is option A, a double reflection
Okay, so let's make a sample space where in one section, one toy is the leader of each group for as many possibilities as there are.
T K E
T E K
K E T
K T E
E T K
E K T
There are 6 possibilities, and so each arrangement has a 1 over 6 chance of being used.
Okay this one is pretty easy.
Here are the steps:
1. we know that 70 - 79 = 5 students 80 - 89 = 7 students 90 - 99 = 2 students.
2. ADD
3. 5 + 7 + 2= 14
4. ANSWER
A: D there are 14 students scored 70 and above!
Hope this helped! (tell me if i'm wrong or right please!!!)
