Question

determine whether the relation shown in the table is a direct variation, an inverse or neither. if it is a direct, write an equation for function.

Answer 1

Direct variation: \(y =kx\)
Indirect variation: \(y = k/x\)

Answer 2

Is there a number \(k\) you can multiply by each value of \(x\) in the table to get the corresponding value of \(y\)? If so, it is direct variation.

Answer 3

If not direct variation, is there a number \(k\) which you can divide by each value of \(x\) in the table to get the corresponding value of \(y\)? If so, it is indirect variation.

Answer 4

Figure out if it is one of those two cases and report back. We'll do whatever other steps might be needed at that point.

Answer 5

what is on there is everything

Answer 6

Yes, all the information you need is there, combined with my instructions here.

Answer 7

?

Answer 8

Do you understand anything of what I've said here? Just trying to figure out where to start explaining, not judging...

Answer 9

no. i dont understand anything.

Answer 10

Okay. Direct variation between x and y means that the data obeys the rule \[y = kx\] where \(k\) is a constant called the constant of variation. For any of \(x\), just multiply it by \(k\) and out comes \(y\).

Answer 11

Now, take the first column in your table. What are the values of \(x\) and \(y\)?

Answer 12

10, and 1/2

Answer 13

\(x = 10\) and \(y = 1/2\)

Okay, using our formula:\[y = kx\]and our known values of \(x,y\):
\[y = kx\]\[\frac{1}{2} = k*10\]Can you solve that for \(k\)?

Answer 14

Hello?

Answer 15

sooo im confused what is the answer?

Answer 16

Can you solve \[\frac{1}{2} = k*10\] for \(k\)?

Answer 17

Did you expect me to work the problem for you while you went away? :-)

Answer 18

...sorta but not entirely sorry .

Answer 19

Okay, so, for the last time, can you solve \[\frac{1}{2} = k*10\] for \(k\)?

Answer 20

idk what to do.

Answer 21

You want to get \(k\) alone on one side of the equals sign, and a number on the other.

Answer 22

You can multiply, divide, add, subtract, just need to do the same thing to both sides.

Answer 23

If I told you that 10 times my favorite number was 40, could you figure out what my favorite number is?

Answer 24

is it 20?

Answer 25

no.

\[\frac{1}{2} = k*10\]Let's multiply both sides by 2\]\[2*\frac{1}{2} = 2*k*10\]\[1=20k\]Now let's divide both sides by 20\[\frac{1}{20} = k\]

Answer 26

Now we need to decide if the other numbers in the table fit this relationship:\[y=kx\]\[y=\frac{1}{20}x\]

Answer 27

Pick the next pair from the table. What is it? Does it satisfy that equation?

Answer 28

15, and 1/3

Answer 29

Okay, plug \(x=15\) and \(y = 1/3\) into the formula. Does it work?

Answer 30

Does \[\frac{1}{3} = \frac{1}{20}*15\]

Answer 31

no

Answer 32

.125 is what is equals

Answer 33

Okay, so that means the data in this table does not follow the rule \(y = kx\) for a single value of \(k\). On to the next type of variation!

Answer 34

Indirect variation or inverse variation means that \[y = \frac{k}x\]or \[k = xy\]

Take the first pair of numbers in the table. Multiply them together. Make a note of the result. Take the second pair of numbers in the table. Multiply them together. Did you get the same result? If not, it is not indirect variation. If it is, you need to try all of the other pairs in the table to make sure they all work.

Answer 35

not all of them work

Answer 36

Really? What do you get for the 4 results?

Answer 37

1/20, .125, 5, 5

Answer 38

Seriously? Show me the pairs of numbers you multiplied together to get those answers.

Answer 39

20, and 1/4 and 25, and 1/5

Answer 40

was the last 2

Answer 41

yes, those are correct. it's the other two that I'm worried about...

Answer 42

you said the 1st was 1/20. and you didnt say anything when i said the 2nd was .125

Answer 43

No, I didn't say that, I said k=1/20 when we were trying direct variation. I told you for this part:

"Take the first pair of numbers in the table. Multiply them together. Make a note of the result. Take the second pair of numbers in the table. Multiply them together. Did you get the same result? If not, it is not indirect variation. If it is, you need to try all of the other pairs in the table to make sure they all work."

Answer 44

First pair of numbers in table are x=10, y = 1/2
Second pair of numbers in table are x = 15, y = 1/3

Answer 45

oh ok they're all 5

Answer 46

so what is the equation?

Answer 47

@whpalmer4

Answer 48

yes, they all produce the same product. that means \[xy = k\] or \[y = \frac{k}x\]The value of \(k\) is just whatever you got each time you multiplied them!

Answer 49

sooo y=k/x?

Answer 50

that's the form, or \(xy=k\), either one. you do see that they are equivalent, right?

Answer 51

but you need to fill in the value of \(k\) in either case!

Answer 52

no i didnt see that so i can write either of them?

Answer 53

yes. \[y=\frac{k}x\]multiply both sides by \(x\)\[x*y = \frac{k}{\cancel{x}}*\cancel{x}\]\[xy=k\]

Answer 54

ok so its indirect?

Answer 55

yes, indirect variation. \(k=\)?

Answer 56

5

Answer 57

so for all the marbles, can you write the complete equation including the constant of variation, aka \(k\)?

Answer 58

yes

Answer 59

drum roll...

Answer 60

?

Answer 61

I'm waiting for you to write it!

Answer 62

write what?

Answer 63

the complete equation, including the constant of variation, aka \(k\)

Answer 64

indirect y=5/x?

Answer 65

there's no ? in the answer, but yes!

Answer 66

ok cool thanks

Answer 67

Now let's check our work by plugging in some values from the table:

\[x=10, y= 1/2\]\[y = 5/10 = 1/2\checkmark\]\[x=15, y = 1/3\]\[y=5/15 =1/3\checkmark\]Looks good!