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lbvjy [14]
3 years ago
11

Is this relation a function? Justify your answer.

Mathematics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

B

Step-by-step explanation:

Eveey input must have exactly one output, x is always the input and y is the output and there are two y outputs on one x input so it is not a function.

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!

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3. C 67.5 mi
4. A 65mi
5. B $270
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3 years ago
213,415.16 in expanded form
weqwewe [10]
200,000 + 10,000 + 3,000 + 400 + 10 + 6 + 0.1 + 0.06
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4. A survey found that of 26 food restaurants, 13 were pizza parlours, 9 did not sell alcohol, and 7 were pizza
ICE Princess25 [194]

Answer: 16/13

Step-by-step explanation:

Facts:  

Total: 26 restaurants

Pizza parlors: 13

9/13 do not sell alcohol

7/13 do not sell alcohol

Total probability (not serving alcohol): 16/13

8 0
2 years ago
I need help on 1,2,3
vladimir2022 [97]

Answer:

1. 10x + 14

2. 2x + 11y - 7

3. -10x - 11

Step-by-step explanation:

1.

4x+5+7x-3+x+12=\\10x+14

2.

(7x+9y-1)-(5x-2y+6)=\\7x+9y-1-5x+2y-6=\\2x+11y-7

3.

-2(2x+5)-6x-1=\\-4x-10-6x-1=\\-10x-11

4 0
3 years ago
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Pls help ill give brainliest
Ymorist [56]

Answer:

\boxed {\boxed {\sf (5g^2+2)(2g-5)}}

Step-by-step explanation:

We are given the expression:

10g^3-25g^2+4g-10

There are no common factors between the four numbers, however the first two have a common factor and the last two do too. Therefore, we can factor by grouping.

Group the first two terms and the last two.

(10g^3-25g^2)+(4g-10)

Find the greatest common factor (GCF) of the first group. It is 5g². Factor it out of the first group. You can do this by dividing both terms by the GCF.

  • 10g³/5g²= 2g
  • -25g²/5g² = -5

5g^2 (2g-5) + (4g-10)

Repeat with the second group. The GCF is 2.

  • 4g/2 = 2g
  • -10/2= -5

5g^2 (2g-5) + 2(2g-5)

There is another GCF: 2g-5. We can factor this out of both terms.

  • 5g²(2g-5)/2g-5= 5g²
  • 2(2g-5)/ 2g-5=2

(5g^2+2)(2g-5)

This cannot be factored further, so it is the answer.

4 0
3 years ago
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