Question #284929

 Use inductive reasoning to make a conjecture about a rule that relates the number you selected to final answer. Try to prove your conjecture by using deductive reasoning.


1. Pick a number


Double it


Subtract 20 from the answer


Divide by 2


Subtract the original number Result:


2. Pick a number Multiply it by 9


Add 21


Divide by 3


Subtract three times the original number Result:


3. Pick a number


Add 6


Multiply the answer by 9


Divide the answer by 3


Subtract 3 times the original number Result




1
Expert's answer
2022-01-06T12:02:34-0500

1.

Let the number be 1.1.

Double it

21=22\cdot1=2

Subtract 20 from the answer


220=182-20=-18

Divide by 2


182=9\dfrac{-18}{2}=-9

Subtract the original number


91=10-9-1=-10

Result:

10-10

Let the number be 13.-\dfrac{1}{3}.

Double it

2(13)=232\cdot(-\dfrac{1}{3})=-\dfrac{2}{3}

Subtract 20 from the answer


2320=623-\dfrac{2}{3}-20=-\dfrac{62}{3}

Divide by 2


6232=313\dfrac{-\dfrac{62}{3}}{2}=-\dfrac{31}{3}

Subtract the original number


313(13)=10-\dfrac{31}{3}-(-\dfrac{1}{3})=-10

Result:

10-10

Let the number be 2.\sqrt{2}.

Double it

2(2)=222\cdot(\sqrt{2})=2\sqrt{2}

Subtract 20 from the answer


2220=22202\sqrt{2}-20=2\sqrt{2}-20

Divide by 2


22202=210\dfrac{2\sqrt{2}-20}{2}=\sqrt{2}-10

Subtract the original number


2102=10\sqrt{2}-10-\sqrt{2}=-10

Result:

10-10

A conjecture: the result is always 10.-10.


Take arbitrary number cR.c\in\R.

Double it

2(c)=2c2\cdot(c)=2c

Subtract 20 from the answer


2c20=2c202c-20=2c-20

Divide by 2


2c202=c10\dfrac{2c-20}{2}=c-10

Subtract the original number


c10c=10c-10-c=-10

Result:

10-10


We see that the result does not depend on the taken number.

Therefore we proved that the result is always 10-10 regardless of the number we pick.


2.

Let the number be 1.1.

Multiply it by 9

19=91\cdot9=9

Add 21


9+21=309+21=30

Divide by 3


303=10\dfrac{30}{3}=10

Subtract three times the original number


103(1)=710-3(1)=7

Result:


77

Let the number be 12.\dfrac{1}{2}.

Multiply it by 9

129=92\dfrac{1}{2}\cdot9=\dfrac{9}{2}

Add 21


92+21=512\dfrac{9}{2}+21=\dfrac{51}{2}

Divide by 3


5123=172\dfrac{\dfrac{51}{2}}{3}=\dfrac{17}{2}

Subtract three times the original number


1723(12)=7\dfrac{17}{2}-3(\dfrac{1}{2})=7

Result:


77

Let the number be 2.\sqrt{2}.

Multiply it by 9

29=92\sqrt{2}\cdot9=9\sqrt{2}

Add 21


92+21=92+219\sqrt{2}+21=9\sqrt{2}+21

Divide by 3


92+213=32+7\dfrac{9\sqrt{2}+21}{3}=3\sqrt{2}+7

Subtract three times the original number


32+73(2)=73\sqrt{2}+7-3(\sqrt{2})=7

Result:


77

A conjecture: the result is always 7.7.


Take arbitrary number cR.c\in\R.

Multiply it by 9

c9=9cc\cdot9=9c

Add 21


9c+21=9c+219c+21=9c+21

Divide by 3


9c+213=3c+7\dfrac{9c+21}{3}=3c+7

Subtract three times the original number


3c+73c=73c+7-3c=7

Result:

77


We see that the result does not depend on the taken number.

Therefore we proved that the result is always 77 regardless of number we pick.


3.

Let the number be 1.1.

Add 6


1+6=71+6=7


Multiply the answer by 9

79=637\cdot9=63

Divide the answer by 3


633=21\dfrac{63}{3}=21

Subtract 3 times the original number


213(1)=1821-3(1)=18

Result:


1818



Let the number be 12.-\dfrac{1}{2}.

Add 6


12+6=112-\dfrac{1}{2}+6=\dfrac{11}{2}


Multiply the answer by 9

1129=992\dfrac{11}{2}\cdot9=\dfrac{99}{2}

Divide the answer by 3


9923=332\dfrac{\dfrac{99}{2}}{3}=\dfrac{33}{2}

Subtract 3 times the original number


3323(12)=18\dfrac{33}{2}-3(-\dfrac{1}{2})=18

Result:


1818



Let the number is 2.\sqrt{2}.

Add 6


2+6=2+6\sqrt{2}+6=\sqrt{2}+6


Multiply the answer by 9

(2+6)9=92+54(\sqrt{2}+6)\cdot9=9\sqrt{2}+54

Divide the answer by 3


92+543=32+18\dfrac{9\sqrt{2}+54}{3}=3\sqrt{2}+18

Subtract 3 times the original number


32+183(2)=183\sqrt{2}+18-3(\sqrt{2})=18

Result:


1818


A conjecture: the result is always 18.18.


Take arbitrary number cR.c\in\R.

Add 6


c+6=c+6c+6=c+6


Multiply the answer by 9

(c+6)9=9c+54(c+6)\cdot9=9c+54

Divide the answer by 3


9c+543=3c+18\dfrac{9c+54}{3}=3c+18

Subtract 3 times the original number


3c+183(c)=183c+18-3(c)=18

Result:


1818


We see that the result does not depend on the taken number.

Therefore we proved that the result is always 1818 regardless of number we pick.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS