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3 years ago

Describe and correct the error in the statement about the relation shown in the table. Select all that apply.

Mathematics

2 answers:

Olegator [25]3 years ago

6 0

Answer:

<u>According to provided table we have:</u>

x- values = domain = 1, 2, 3, 4, 5
y- values = range = 6, 7, 8, 9

We know that a function is a such relationship where each input has exactly one output.

<u>Now, the answer choices:</u>

A) The relation is not a function because the output 6 has two inputs.

Incorrect, a function can have multiple inputs with same output.

B) The relation is not a function because one output is paired with multiple outputs.

Incorrect, there is no such relationship according to given table.

C) The given list is the domain of the function. The range is 6, 7, 8, 9.

The list of domain is not given but as shown above it is 1, 2, 3, 4, 5. The range is correct.

This is a correct choice.

D) The given list is missing elements of the range. The range is 1, 2, 3, 4, 5, 6, 7, 8, 9

Incorrect, this is a combination of the domain and the range.

Leno4ka [110]3 years ago

3 0

The answer is C
the domain is the x values which would be {1,2,3,4,5} and the range is the y values which would be {6,7,8,9}
if this helped please give brainliest!

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Select the correct answer.

Scrat [10]

Answer: C Roger’s sister has been working as a skydiving instructor for the last six months.

Step-by-step explanation: really the most life risking or the only thing on that list that could need life insurance.

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3 years ago

Given that sec0=-37/12, what is the value of cot0, for pi/2<0

Colt1911 [192]

Answer:

2nd option

Step-by-step explanation:

Using the identities

tan²θ + 1 = sec²θ

cotθ = $\frac{1}{tan0}$

Given

secθ = - $\frac{37}{12}$ , then

tan²θ + 1 = (- $\frac{37}{12}$ )² = $\frac{1369}{144}$ ( subtract 1 from both sides )

tan²θ = $\frac{1225}{144}$ ( take the square root of both sides )

Since $\frac{\pi }{2}$ < θ < π and tanθ < 0 , then

tanθ = - $\frac{35}{12}$

Then

cotθ = $\frac{1}{-\frac{35}{12} }$ = - $\frac{12}{35}$

5 0

3 years ago

Determine whether each of these functions from Z to Z is one-to-one. a) f(n) = n - 1 b) f(n) = n2 + 1 c) f(n) = nº d) f(n) = [n/

sesenic [268]

Answer: The correct option is

(a) f(n) = n - 1.

Step-by-step explanation: We are given to determine whether the given functions are one-to-one or not.

We know that a function y = f(x) is one-to-one if and only if

f(x) = f(y) ⇔ x = y.

That is, any two distinct elements cannot have the same image.

(a) The given function is

$f(n)=n-1.$

Let us consider that

$f(n_1)=f(n_2)\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow n_1=n_2.$

Similarly,

$n_1=n_2\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow f(n_1)=f(n_2).$

So, this function is one-to-one.

(b) The given function is

$f(n)=n^2+1.$

Let us consider that

$f(n_1)=f(n_2)\\\\\Rightarrow n_1^2+1=n_2^2+1\\\\\Rightarrow n_1^2=n_2^2\\\\\Rightarrow n_1=\pm n_2.$

That is, there may be two unequal elements having same image.

For example, f(-1)=(-1)²+1=1+1=2, f(1)=(1)²+1=1+1=2.

It implies that f(-1)=f(1) but 1 ≠ -1.

So, the given function is not one-to-one.

(c) The given function is

$f(n)=n^0.$

Here, the image of all the elements is 1.

For example, $f(2)=2^0=1,~~f(3)=3^0=1.$

f(2)=f(3) but 2≠3.

So, more than one element is having the same image and so the function cannot be one-to-one.

(d) The given function is

$f(n)=\left[\dfrac{n}{2}\right].$

Here, we see that

$f(2)=\left[\dfrac{2}{2}\right]=[1]=1,\\\\\\f(3)=\left[\dfrac{3}{2}\right]=[1.5]=1.$

So, f(2)=f(3) but 2≠3.

So, the given function is not one-to-one.

Thus, the correct option is (a).

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iren2701 [21]

B = 2a/ch that's your answer

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Igoryamba

Answer: the answer is d

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