I believe they are all false
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
#SPJ1
Complete Question:
State any restrictions on the variables 23/7y
Answer:
9676
Step-by-step explanation:
Answer:
2sin(3θ) - √3 = 0
Step-by-step explanation:
sin(3θ) = √3/2
3θ = π/3 + 2kπ or 2π/3 + 2kπ, k = 0, ±1, ±2, ±3,...
θ = π/9 + 2kπ/3, 2π/9 + 2kπ/3
If k = 0, we get θ = π/9, 2π/9
If k = 1, we get θ = 7π/9, 8π/9
If k = 2, we get θ = 13π/9, 14π/9
Other values of k give values of θ lying outside of the interval [0, 2π).
Answer:
43b. (-1, 2)
43a. (1, -3)
36b. Yes: (0,9)
Step-by-step explanation:
<u>43b</u>. x+y=1
-x+y=3
----------------------
2y=4 > y=2
2+x=1 > x=-1
check:
2(-1)+2=0
0=0
<u>43a</u>. can you figure this one out by yourself and then check it? If you can't I will show my work. The answer is above. :)
<u>36b</u>. y-(0)^2=9
y=9
