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3 years ago

What is the domain of the given relation {(1,3), (0,4), (2,1)}?

Mathematics

1 answer:

myrzilka [38]3 years ago

4 0

Answer:

The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements "used" by the relation or function constitute the range.

Step-by-step explanation:

I'll just list the x-values for the domain and the y-values for the range: domain: {–3, –2, –1, 0, 1, 2}. range: {5}

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Consider the expression √(√625).

const2013 [10]

Answer:

Step-by-step explanation:

Square Root - Finding a number that multiplies itself twice into the number within the square root

With this meaning, we need to find a number that multiplies itself into 625.

Finding calculations using exponents will help,

25^2 is equal to 625. Therefore $\sqrt{25}$

We are not finished, as there is another square root right after, empowering the parenthesis, so what number multiplies itself equals 25?

This would be 5, therefore $\sqrt{(\sqrt{625}) }$ is equal to 5.

6 0

3 years ago

What are the like terms in the expression below? 3x + 8 + 3y + 8x

Lorico [155]

The like terms would be 3x and 8x

4 0

3 years ago

What would the first step be in solving the inequality: -y (less than or greater than sign) 136

Alex73 [517]

"-y (less than or greater than sign) 136" translates into two separate inequalities:

-y < 136 and -y > 136.

You want to solve for y in each case. To do this, multiply each inequality by "-1" and then immediately change the direction of the inequality sign.

For example: -1(-y < 136) => y > -136.

5 0

3 years ago

-4k^2-8k-3 = -3-5k^2

mel-nik [20]

$-4k^2-8k-3 = -3-5k^2 \\ \\ -4k^2 - 8k = -5k^2 \\ \\ 4k^2 + 8k - 5k^2 = 0 \\ \\ -k^2 + 8k = 0 \\ \\ 8k - k^2 = 0 \\ \\ k(8 - k) = 0 \\ \\ k = 0,8 \\ \\$

The final result is: k = 0, 8.

8 0

3 years ago

Read 2 more answers

X < 36 and x > -36 x < 36 or x>-36 x< 36 or x > 0 x < 36 and x > 0

hjlf

Answer:

D,x < 36 and x > 0, negative number cannot be in sq.root

3 0

3 years ago