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shepuryov [24]
3 years ago
5

2. State the domain for the following ordered pairs: { (4, -2), (6, -7), (4, 3), (-8, 1), (-3, 5) } *

Mathematics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

(4,6,4,-8,-3)

Step-by-step explanation:

Domain means of function is the set of all possible inputs for function

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Please answer this correctly
meriva

Answer:

33.3%

Step-by-step explanation:

The numbers greater than 6 from the spinner are 7 and 8.

2 numbers out of total 6 numbers.

2/6 = 1/3

= 0.333

= 33.3%

4 0
3 years ago
Equation which i dont like math not no brainly my question 3 (k-12)=k-30
vaieri [72.5K]
3 (k-12) = k-30

3k - 36 = k - 30

3k - k = -30 + 36

2k = 6

k = 3
7 0
3 years ago
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Which of the following modifications to the list of assets and liabilities below would result in an increase in net worth?
jolli1 [7]

The modifications made to the list of assets and liabilities that would result in an increase in net worth is Option D. Net worth is defined as the value of all the non-financial and financial assets owned by an individual or an institution.

All of the above (Putting $100 in savings, Paying $100 on credit cards and Getting paid $100)

5 0
3 years ago
Read 2 more answers
Arithmetic sequence
KonstantinChe [14]

Answer: Arithmetic Sequence

Step-by-step explanation:

An Arithmetic Sequence is a sequence where each term increases by adding/subtracting some constant k. This is in contrast to a geometric sequence where each term increases by dividing/multiplying some constant k.

5 0
2 years ago
Evaluate $\dfrac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}$.
Pavlova-9 [17]
<h3>Answer:  4</h3>

========================================================

Work Shown:

\frac{2\sqrt{72}}{\sqrt{8}+\sqrt{2}}\\\\\frac{2\sqrt{36*2}}{\sqrt{4*2}+\sqrt{2}}\\\\\frac{2\sqrt{36}*\sqrt{2}}{\sqrt{4}*\sqrt{2}+\sqrt{2}}\\\\\frac{2*6*\sqrt{2}}{2*\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{2\sqrt{2}+\sqrt{2}}\\\\\frac{12\sqrt{2}}{3\sqrt{2}}\\\\\frac{12}{3}\\\\4

Note in step 2, I factored each number in the square root to pull out the largest perfect square factor. From there, I used the rule that \sqrt{A*B} = \sqrt{A}*\sqrt{B} to break up the roots.

8 0
2 years ago
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