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bearhunter [10]
3 years ago
8

Which statement is false? A: Some integers are irrational. B: Some integers are whole numbers. C: Some rational numbers are inte

gers. D: Some real numbers are irrational.
Mathematics
2 answers:
djverab [1.8K]3 years ago
6 0

Answer:

A

Step-by-step explanation:

A is false because irrational numbers can not be integers.

Illusion [34]3 years ago
3 0
A: some integers are irrational
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Vsevolod [243]

Answer:

x =  1, y = 3

x = -5, y = -3

Step-by-step explanation:

Point Form:

(1,3) , (-5, -3)

4 0
2 years ago
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Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:
Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2

(4+1)^2=(4-1)^2+(4)^2

(5)^2=(3)^2+(4)^2

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2

(9+1)^2=(9-1)^2+(6)^2

(10)^2=(8)^2+(6)^2

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2

(9+4)^2=(9-4)^2+(12)^2

(13)^2=(5)^2+(12)^2

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2

(16+1)^2=(16-1)^2+(8)^2

(17)^2=(15)^2+(8)^2

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2

(16+4)^2=(16-4)^2+(16)^2

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We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2

(16+9)^2=(16-9)^2+(24)^2

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We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2

(25+1)^2=(25-1)^2+(10)^2

(26)^2=(24)^2+(10)^2

So (10,24,26) is another.

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(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2

(25+4)^2=(25-4)^2+(20)^2

(29)^2=(21)^2+(20)^2

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2

(25+9)^2=(25-9)^2+(30)^2

(34)^2=(16)^2+(30)^2

Another Pythagorean Triple is (16,30,34).

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(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2

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1 year ago
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Sveta_85 [38]

Answer: 4 servings

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3 years ago
5 points
zubka84 [21]

Answer:

d. $28.70

Step-by-step explanation:

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