Question

I need help on task 2 pg 67 algebra 1

Answer 1

heres the problem:

Answer 2

@Mokeira

Answer 3

The top table, based on what it has given you, you need to find the equation. It is linear, so you need to use y=mx+b. That equation, should help you find when x+5, 6, and 7. Try that, if you want more explaination, I will give it. :) It's important to try to work these out yourself as much as you can.

Answer 4

its the side table that they are talking about @dezmwas24

Answer 5

yeah...finding the equation will make it easier

Answer 6

whats a gp @Mokeira

Answer 7

i mean AP...sorry!!!

Answer 8

GP- geometric progression
AP-Arithmetic progression

Answer 9

I could be wrong, but it might be going up exponentially. What have you covered in the chapter?

Answer 10

I thought it was going up by x2

Answer 11

i thought we were adding multiples of 2

Answer 12

yah that's what I meant

Answer 13

so what would y be when x=5

Answer 14

am I on the right track

Answer 15

i think you are..i would do the exact same thing

Answer 16

ok so now I simplify the numbers by adding

Answer 17

Your reasoning is right, so yeah, you're on the right track. You still have to right an equation, right?

Answer 18

yes...that looks like part (b)

Answer 19

we still need an equation :( lol....

Answer 20

yes @dezmwas24 but first I have to add the y values

Answer 21

one second

Answer 22

Right.

Answer 23

hmm....the equation is also super easy to get...shouldnt be hard. I was shying away from trying

Answer 24

so now I have this

Answer 25

have you added already?

Answer 26

yes is that right

Answer 27

Do you know how to find the equation?

Answer 28

no

Answer 29

i think you have sent the wrong one...but i am sure you are right...so now let us get the equation

Answer 30

sorry

Answer 31

you are right...how do we get the equation @dezmwas24

Answer 32

??

Answer 33

@dezmwas24

Answer 34

Sorry.

@247 Have you written an exponential equation before?

Answer 35

I don't think so what is that

Answer 36

ohh wait yes I have

Answer 37

Does this formula look familiar?

Answer 38

yes

Answer 39

Your y values, if you noticed, they don't go up by the same amount every time. It goes up by 2, then 4, then 8, and so on. I think that's what makes it an exponential function.

Answer 40

ok but I have no idea how to find the equation

Answer 41

@JoannaBlackwelder @Mashy @ganeshie8

Answer 42

@Mokeira

Answer 43

I am so sorry. It has been a bit since I've done this, but I will try to figure it out.

Answer 44

thank you

Answer 45

@Gman344 @J.F @kirbykirby @inkyvoyd

Answer 46

\[\Large y = \sum_{x=1}^n2^{x-1} \]

Answer 47

You can use the formula for the sum of a geometric progression to find the value for x=15, or just sum it by hand

Answer 48

@kirbykirby That's for the sum, not for an exponential equation.

Answer 49

but \(y\) is finding a sum of terms
when x = 3, y is \(2^0+2^1+2^2\)

Answer 50

y is doing a sum of exponential functions if you want to view it that way

Answer 51

So that equation you wrote earlier was the answer? I think I get it now.

Answer 52

@247 Did you see what kirbykirby put?

Answer 53

yes you can use that, but also a more efficient formula is using the formula for the sum of a geometric progression:
\[\large \sum_{x=1}^nar^{x-1}=a\left( \frac{1-r^n}{1-r}\right) \] In this specific example we have
\[\large \sum_{x=1}^n(1)(2)^{x-1}= \frac{1-2^n}{1-2} \]

Answer 54

For part (b), your list looks good.
There is an interesting pattern you may not notice in your numbers
x y y+1
1 1 2
2 3 4
3 7 8
4 15 16
5 31 32
6 63 64
7 127 128

do you see that y+1 is a power of two?
we can use that to find a formula to predict y when we know x

Answer 55

to make it more clear
x y y+1 y+1
1 1 2 2
2 3 4 2*2
3 7 8 2*2*2
4 15 16 2*2*2*2
5 31 32 2*2*2*2*2
6 63 64 2*2*2*2*2*2
7 127 128 2*2*2*2*2*2*2

Answer 56

the idea is find y+1 from x (knowing it is a power of 2), then subtract 1 to get y

Answer 57

That will be your answer for part C

Answer 58

so @phi what will y be when x=15

Answer 59

to find y when x is 15, you first need the formula
Did you figure out the formula?

Answer 60

no I don't know it @phi

Answer 61

This is from yesterday

x y y+1 y+1
1 1 2 2
2 3 4 2*2
3 7 8 2*2*2
4 15 16 2*2*2*2
5 31 32 2*2*2*2*2
6 63 64 2*2*2*2*2*2
7 127 128 2*2*2*2*2*2*2


any idea what the rule is for finding y+1 when x is 8 ?

Answer 62

no

Answer 63

Look at each x value, and then at how to compute y+1 for that x value
what is the pattern ?

Answer 64

I don't understand @phi

Answer 65

are you saying you can't find a pattern in this:
x y+1
1 2
2 2*2
3 2*2*2
4 2*2*2*2
5 2*2*2*2*2
6 2*2*2*2*2*2
7 2*2*2*2*2*2*2

Answer 66

I don't have those numbers I have these @phi

Answer 67

@Mokeira

Answer 68

Yes, I know what numbers you have.

Put that aside for the moment. Can you find the pattern in
x y+1
1 2
2 2*2
3 2*2*2
4 2*2*2*2
5 2*2*2*2*2
6 2*2*2*2*2*2
7 2*2*2*2*2*2*2

Answer 69

yes you add a *2 everytime

Answer 70

be more specific. for example, when x=3, what is the rule for (y+1) ?

Answer 71

This is close to being answered, if you try a little...

Answer 72

\[2^{3}\] @phi

Answer 73

yes, and if we generalize to x=n, y+1 = 2^n

now back to your problem
x y y+1 y+1 y+1
1 1 2 2 2^1
2 3 4 2*2 2^2
3 7 8 2*2*2 2^3
4 15 16 2*2*2*2 2^4
5 31 32 2*2*2*2*2 2^5
6 63 64 2*2*2*2*2*2 2^6
7 127 128 2*2*2*2*2*2*2 2^7

your list of numbers is the list labeled 'y' , the second column

how do we change y+1 to get y ?

Answer 74

I don't get where your getting these numbers @phi

Answer 75

*2

Answer 76

I noticed if we add 1 to your numbers (labeled "y" in the table) you get the numbers in the y+1 column, which (if you know your powers of 2) are powers of 2
It's easy to find the rule to find the numbers in the y+1 column
if x is n , the number in the y+1 column is 2^n
to get your number, subtract 1

so use
2^n - 1
as your rule.

Answer 77

ok