Question

does anyone know how to do this one. write the difference in expanded form 893- 514 using exponents

Answer 1

not sure how to do this

Answer 2

well, I'm not sure what they mean... but it sounds like
893 - 514 = 379

now we can say that we can EXPAND 379 to be a sum
like
81+36+25 + 169 + 49 + 19

Answer 3

\(\bf 81+36+25 + 169 + 49 + 19 \implies 9^2+6^2+5^2+13^2+7^2+19\)

Answer 4

I think she wants me to write it something like this https://angel.spcollege.edu/AngelUploads/QuestionData/37d2f850-c4fd-4c48-9749-8fc615336511/36232G36223323412361.png# {0af42309-0f91-4d18-921b-c02083637aef}

Answer 5

need to write it like that and dont remember how

Answer 6

ok, so we know that the difference is 379

so can you provide.... say 3 numbers, that added together will give 379?

1st number must end with "00", 2nd number must end with "0" and the last one will be what's remaining of the 379

Answer 7

lost

Answer 8

can you think of a number that ends with "00" <--- zero zero

Answer 9

lemm show you what the picture there stand for

Answer 10

10

Answer 11

100

Answer 12

\(\large \begin{array}{llll}
(3\times 10^2)+(8\times 10^1)+(1\times 10^0)\\ \quad \\ \quad \\
\begin{array}{llll}
10^0 & = & 1\\
10^1 & = & 10\\
10^2 & = & 100\\
10^3 & = & 1000\\
10^4 & = & 10000\\
...
\end{array}\\
\textit{as many zeros as the exponent shows}\\
(3\times 10^2)\implies 3 \times 100 = 300\\
(8\times 10^1)\implies 8 \times 10 = 80\\
(1\times 10^0)\implies 1 \times 1 = 1\\ \quad \\ \quad \\
(3\times 10^2)+(8\times 10^1)+(1\times 10^0) = 300+80+1



\end{array}\)

Answer 13

so 100
ok... can you give me say.... another ending in "0" <- zero

Answer 14

10\[10^2\]

Answer 15

Looking at the example from earlier post, I think I get what they want you to practice. Remember that \[1=1*10^{0}.\]. Here is the middle term:
\[-514= -1 * ( 5*10^{2} + 1*10^{1} + 4*10^{0} ) = -5*10^{2} - 1*10^{1} - 4*10^{0} \]
Then group exponents together and simplify. Example: \[8*10^{2} - 5*10^{2} = (8-5)*10^{2} = 3*10^{2} ==> 800 -500=300\]

Does this help?
Putting the whole break down and steps into the equation writer would take a while...

Answer 16

yes

Answer 17

so for instance this would be Simplify the expansion https://angel.spcollege.edu/AngelUploads/QuestionData/03263c60-f44c-4b6d-934d-07422994703e/221132662322HR573221.png# {018270b7-a112-45a1-8477-f874d215b8a5}

Answer 18

10, 795

Answer 19

hmm, ... ok, let's use 10
so we have 100 and 10

the difference is 379

100 + 10 = 110
379 - 110 = 269

so our numbers will be 100 + 10 + 269

\(\bf 100 + 10 + 269\\ \quad \\ \quad \\
100 = 10^2 \implies 1 \times 10^2\\
10 = 10^1 \implies 1 \times 10^1\\
269 = 269 \implies 269 \times 10^0\\ \quad \\ \quad \\
(1 \times 10^2)+(1 \times 10^1)+(269 \times 10^0)\)

Answer 20

ok

Answer 21

Not quite. We count in base-10. Think of the exponent as the number of zeros that follow the multiplier. your example is 1000 + 700 + 90 + 5 = 1795.

Answer 22

right

Answer 23

I am lost now

Answer 24

Sorry... simpler example. Maybe I can get a simpler one in equation format. Give me two three-digit number to subtract in exponential format.

Answer 25

514-345

Answer 26

Ah! Good example... give me a minute.

Answer 27

ok

Answer 28

\[514-345 => 5*10^2 + 1*10^1 + 4*10^0 - ( 3*10^2 + 4*10^1 + 5*10^0 ) \]

Answer 29

Working on next transform. (That's the most the equation writer could handle).

Answer 30

\[5*10^2 +1*10^1+4*10^0 - 3*10^2 - 4*10^1 - 5*10^0\]

Answer 31

Combine terms by exponent value: \[(5-3)*10^2 + (1-4)*10^1 + (4-5)*10^0\]

Answer 32

ok

Answer 33

no that seems easier

Answer 34

\[2*10^2 - 3*10^1 - 1*10^0\]

Answer 35

Looking at this now: You can see 200-30-1 = 169 , which is the answer you get from a calculator. Do you need to complete the exercise just with exponents? That is a little more work.

Answer 36

yes, I know that part

Answer 37

OK. Did this help?

Answer 38

Here is the final exponent reduction "just for fun" : \[(2 - .3 - .01)*10^2 = 1.69*10^2 = 169\]

Answer 39

yes

Answer 40

since you explained it , it seems much easier now

Answer 41

I was lost from the start

Answer 42

WooHoo!!! I helped someone! ;-)
I do these things at work and just get puzzled looks.
Work on understanding the details of the last transformation. It's the key to a lot of engineering and chemistry calculations.

Answer 43

yes :)))

Answer 44

I know i think i have it down, I wish I could understand my anatomy classes like i understand this