Question

HElp!

Answer 1

@pitamar

Answer 2

help me!!!!!!! http://openstudy.com/study#/updates/54ecbbbbe4b0dc552b63e7c0

Answer 3

no pit help MEEEEE

Answer 4

@welshfella

Answer 5

i need help my class is about to be over :,(

Answer 6

Ok. Let's start with finding the formula for this.
We know that \(a_0 = 40\), right? And \(a_1 = 20\) then \(a_2 = 10\), so we can see that every time the quantity is multiplied by \(\frac{1}{2}\).
So we could say that:
$$
a_n = 40 \cdot \Big(\frac{1}{2}\Big)^n
$$is it clear so far?

Answer 7

I can't read your document

Answer 8

yes that is clear

Answer 9

So we have to find what term is that 0.078125.
We know that:
$$
40 \cdot \Big(\frac{1}{2}\Big)^n = 0.078125
$$Can you solve it?

Answer 10

honestly i cant do this one

Answer 11

Well, let's start by dividing both sides by 40. what do we get?

Answer 12

@welshfella I attach a picture of the problem =)

Answer 13

how did you make it small like that?

Answer 14

I cropped it in a graphic program. you can do it in Paint too:
http://www.wikihow.com/Crop-an-Image-with-Microsoft-Paint

Answer 15

ok so we know it has to be 79.

Answer 16

what is 0.078125 / 40?

Answer 17

let me find a cal one sec

Answer 18

https://www.google.co.il/search?q=calc
=)

Answer 19

0.00195312

Answer 20

Well I get 0.001953125
Another 5 at the end. but ye, good job.
So we have:
$$
\Big(\frac{1}{2}\Big)^n = 0.001953125
$$Right?

Answer 21

yes

Answer 22

You know that \(\Big(\frac{a}{b}\Big)^n = \frac{a^n}{b^n}\) ?

Answer 23

i didnt but gotcha

Answer 24

so \(\Big(\frac{1}{2}\Big)^n = \frac{1}{2^n}\)
Clear?

Answer 25

ok i understand that so yes clear

Answer 26

So our equation becomes
$$
\frac{1}{2^n} = 0.001953125
$$And now we can divide 1 by both sides. Means:
$$
\frac{1}{\frac{1}{2^n}} = \frac{1}{0.001953125}
$$And that becomes:
$$
2^n = \frac{1}{0.001953125}
$$Clear so far?

Answer 27

Yes!

Answer 28

ok, so calculate \(\frac{1}{0.001953125}\)

Answer 29

512

Answer 30

Correct sir

Answer 31

So
$$2^n = 512$$
We have to find what \(n\) will make this true.
Do you know the answer?

Answer 32

79.9

Answer 33

Nope
We can just make a list:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
....
Can you go keep it going until you get 512?

Answer 34

yeah but isnt there a easier way to do this

Answer 35

Yes, using logarithms. but I have a feeling you don't know those and I wanna keep it simple.

Answer 36

ok so whats next teach

Answer 37

You still have to find me what 2^n = 512
Keep the list going:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
....

Each time multiply the last value by 2 and add to the list, until you hit 512. Don't worry it's not long.

Answer 38

ok one sec

Answer 39

ok i got it 20012

Answer 40

actually i wrote it wrong

Answer 41

O.o

Answer 42

i got 1392

Answer 43

heh, let's do it together. If 2^4 is 16. then 2^5 is twice that. how much is 2^5 then?

Answer 44

26464

Answer 45

16*2 = ?

Answer 46

O.o

Answer 47

32

Answer 48

right, so 2^5 = 32.
2^6 = 2^5 * 2 = 32 * 2 = ?

Answer 49

16

Answer 50

32*2 = ?

Answer 51

32*2 = (30 + 2)*2 = 30*2 + 2*2 = 60 + 4 = 64
Get it?

Answer 52

yes but how is that even close to the choices?

Answer 53

but yes i get it

Answer 54

Leave the choices. we're half way through. we have to determine 'what' we have to sum before we sum it. so 2^6 = 64 * 2 = ?

Answer 55

oh ok gotcha so 16

Answer 56

64*2 is not 16 =|
use the calculator 64*2 = ?

Answer 57

128

Answer 58

Right. So 2^7 = 128 . So 2^8 = 128*2 = ?

Answer 59

256

Answer 60

Right, 2^9 = ?

Answer 61

18

Answer 62

it's not * it's ^
2^9 = 128 * 2 = ?

Answer 63

ok so then what?

Answer 64

oops
2^9 = 256 * 2 = ?
Calculate and see...

Answer 65

512

Answer 66

Right. So we found 2^9 = 512
We found \(n\) in
$$
2^n = 512 \\
2^9 = 512 \\
n = 9
$$understand?

Answer 67

yes i do

Answer 68

So now we can say:
$$
2^9 = 512 \\
\frac{1}{2^9} = \frac{1}{512} \\
\Big(\frac{1}{2}\Big)^9 = 0.001953125
$$We found what that 0.001953125 is (1/2)^9
Get it?

Answer 69

ok one sec

Answer 70

i do get it but i mis read it srry

Answer 71

Now by multiplying both sides by the 40 we divided we get:
$$
40 \cdot \Big(\frac{1}{2}\Big)^9 = 40 \cdot 0.001953125 \\
40 \cdot \Big(\frac{1}{2}\Big)^9 = 0.078125\\
$$

Answer 72

ok i am starting to see better what were doing

Answer 73

To remind you, we said:
$$ a_n = 40 \cdot \Big( \frac{1}{2} \Big)^n $$
So for:
$$ 40 \cdot \Big( \frac{1}{2} \Big)^9 = 0.078125 $$
We can say that the \(n\) is 9. So that means \(a_9 = 0.078125\)
Get it?

Answer 74

yes i do get it

Answer 75

it actually makes snese

Answer 76

yes but not good at it

Answer 77

So why we did all this? we did all this because now we know we're summing from \(a_0\) to \(a_9\). So we can write it with sigma notation:
$$\sum_{k=0}^9 a_k = \sum_{k=0}^9 40 \cdot \Big( \frac{1}{2} \Big)^k = 40 \cdot \sum_{k=0}^9 \Big( \frac{1}{2} \Big)^k$$ Are you familiar with this notation?

Answer 78

kinda

Answer 79

Ok, well we have to sum
$$
\sum_{k=0}^9 \Big(\frac{1}{2} \Big)^k
$$Then we can multiply by 40 and get the final result. There is a formula for that:
http://mathworld.wolfram.com/GeometricSeries.html
Look at equation number (6):
$$
S_n \equiv \sum_{k=0}^n r^k = \frac{1 - r^{n+1}}{1-r}
$$What is \(r\) in our case?

Answer 80

ok

Answer 81

i gotcha so far

Answer 82

Ye, but I asked what is \(r\) in our case?
We are trying to sum:
$$
\sum_{k=0}^9 \Big(\frac{1}{2} \Big)^k
$$ So what is the \(r\)?

Answer 83

Compare to the formula

Answer 84

you sub it in right for k

Answer 85

Arm, let's try to use the formula we have to figure out this sum.
Look at the formula again:
$$S_n \equiv \sum_{k=0}^n r^k = \frac{1 - r^{n+1}}{1-r}$$
You see that they are summing \(r^k\)
In our case we have:
$$
\sum_{k=0}^9 \Big(\frac{1}{2} \Big)^k
$$What we have here instead of the \(r\) in the formula?

Answer 86

ok so what is r for then?

Answer 87

the formula i know

Answer 88

Look at the formula, look at what we're trying to sum. In our case we don't have \(r\) we have something else in its place. What is that something-else?

Answer 89

k

Answer 90

or does the E count?

Answer 91

well no. we have \(r^k\) in the formula. in our case we have \( \Big(\frac{1}{2}\Big)^k \). So our \(r\) is \(\frac{1}{2}\). get it?

Answer 92

i do now but before i was guessing cause i didnt know

Answer 93

opk so what next?

Answer 94

Ok there is one more thing we have to compare to our formula:
$$S_n \equiv \sum_{k=0}^n r^k = \frac{1 - r^{n+1}}{1-r}$$
The formula also has \(n\) in it. It sums from k=0 to k=n.
We have
$$\sum_{k=0}^9 \Big(\frac{1}{2}\Big)^k$$
So do we have instead of the \(n\)?

Answer 95

1/2

Answer 96

wow we been on this for two hours :ooo