Which value would not be part of the domain of the inverse of the given relation?

Pls help me with this ans should be well explained

Snezhnost [94]

Let's observe

#1st column

3+2+1+4=10

10th alphabet=J

#2nd

5+9+4+6=24

24th alphabet=X

#3rd

2+1+3+3=9

9 th Alpha bet=I

#4th

8+4+9+5=26

26th Alphabet=Z

#5th

7+3+1+2=13

13 th alphabet=M

Option B is correct

5 0

3 years ago

Can someone please help me

Slav-nsk [51]

Answer:

B

Step-by-step explanation:

The sum of the angles of the triangle add up to 180 degrees.

x and y are equal to each other in this type of the triangle.

4 0

3 years ago

Liz is making 1 dozen ice pops for dessert. She mixes the ingredients together in her kitchen, puts them in the ice-pop mold, an

Troyanec [42]

The answer to your question is 22 degrees Celsius. To figure out the problem, all you do is take the minutes( 58 and 2 and 35 sec.) and convert to seconds(3,480 and 145) Then divide 3,480 by 145 ,which equals 24. Then take 24 and add to the final temperature, which equals 22.
hopes this helps!

3 0

3 years ago

HELP ME RN PLEASEEEE [WILL GIVE BRAINLY TO CORRECT AND BEST EXPLAINED ANSWER]

VMariaS [17]

Answer:

$\boxed {\boxed {\sf d= \sqrt{34}}}$

Step-by-step explanation:

We want to find the distance between two points, so the following formula is used.

$d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Where (x₁, y₁) and (x₂, y₂) are the points we are finding the distance between.

We are given the points (-2, -1) and (3,2). If we match the corresponding value and variable we see that:

x₁= -2
y₁= -1
x₂= 3
y₂= 2

Substitute the values into the formula.

$d= \sqrt {(-2-3)^2+(2--1)^2$

Solve the parentheses.

-2 -3 = -5
2--1 = 2+ 1 = 3

$d= \sqrt{(-5)^2+(3)^2$

Solve the exponents.

(-5)²= -5*-5= 25
(3)²= 3*3=9

$d= \sqrt{25+9}$

Add.

$d= \sqrt{34}$

This radical cannot be simplified, so the distance between the two points is √34 and choice 3 is correct.

6 0

3 years ago

Read 2 more answers

Pythagorean triples are given by the formulas x2 - y2, 2xy, and x2 + y2. Use the formulas for the Pythagorean triples to prove w

Arte-miy333 [17]

Let $k$ be an integer. Suppose there is a triangle with legs of length 16 and $2k+1$ . Then by the Pythagorean theorem, the length of the hypotenuse should be

$\sqrt{16^2+(2k+1)^2}=\sqrt{4k^2+4k+257}$

The formulas for Pythagorean triples say that if the legs are integers, then so must be the hypotenuse, because if $x=16$ and $y=2k+1$ are integers, then so are $x^2-y^2$ , $2xy$ , and $x^2+y^2$ .

However, $4k^2+4k+257$ is not a perfect square trinomial, which means for any integer $k$ , the length of the hypotenuse is not an integer, so such a triangle doesn't exist.

4 0

3 years ago