# (c) (b) (y) (b) (b)

Discussion in 'Math' started by Lightfire, Sep 21, 2011.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello in my assignment notebook, I am confusing with this

$c\cdot b\cdot y\cdot b\cdot b$

We have to express it in exponent

There are three b there but they are part away. what would be the answer in the exponent form?

$cb^{3}y$ Is it like that the answer?

This one also

Theere are two numbers part away. Our teacher said we have to write the answer of numbers, if there is any.

$-5 \cdot x\cdot x\cdot 3\cdot y\cdot y$

$-5 \cdot x\cdot x\cdot 3\cdot y\cdot y =-5x^{2}3y^{2} = -15x^{2}y^{2}$ is the answer like that?

thanks so much

2. ### tgotwalt1158 Member

Feb 28, 2011
111
18
What is the title subject of the assignment chapter?

3. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
What do you mean? The instruction? The assignment was given by our teacher

the instruction.

thank

4. ### tgotwalt1158 Member

Feb 28, 2011
111
18
If the dot between constants and variables is the multiplication symbol, then both of your answers are correct and you are doing right, since in algebra, variable's powers are added when they are multiplied and their co-efficients are multiplied by themselves.

Lightfire likes this.
5. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
You are right, Catapult. When you're multiplying several variables together, it does not make a difference what order they go in. Therefore, C*B*Y*B*B is the same as saying C*Y*B*B*B, which simplifies to C*Y*B^3, or as you said, C*B^3*Y. They all mean the same thing, so you are doing it correctly
Der Strom

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6. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21

Now, I am confused with this.

In our assignment's instruction, we have to simplify. It's negative exponent.

$\frac{s^{-2}}{t}$

My answer is $\frac{s}{t^{2}}$

Is my answer need more simplification.

7. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
With this also. what's the answer.

$\frac{6^{2}x^{0}y^{-3}}{2^{4}}$

$\frac{7^{2}x^{2}y^{-3}}{y^{-4}}$
pls help me

8. ### tgotwalt1158 Member

Feb 28, 2011
111
18
I believe its like this, long time out of practice though. Please confirm before submitting your assignment. No guaranties, just idea!

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9. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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That's not quite right. Since you're raising s to the -2 power, that means that s and the power need to stay together. The power can not just move to the bottom. Does that make sense?

10. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
tgotwalt1158, this is Lightfire's homework, so it would be better to explain how to do it, rather than do it for him. Just something to keep in mind, okay?

11. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
The first thing I see is a lot of numbers and exponents that can be simplified to a single number. For example, let's take the first one--you have 6^2. You can simplify that to a single number. You can do that with most of those numbers with exponents. even x^0 can be simplified to a single number (if you know the rules for that sort of exponent ). I suggest you take it step by step--you can do this
Good luck!
Der Strom

EDIT: Sorry I just answered 3 times in a row

12. ### BillO Distinguished Member

Nov 24, 2008
985
136

Are they asking you to get rid of the negative exponents?

Remember these identities;

$x^{-y}\ \ =\ \ \frac{1}{x^{y}}$

$x^{y}\ \ =\ \ \frac{1}{x^{-y}}$

$x^{o}\ \ =\ \ \1$

Also, something like;

$3^{-3}\ \ =\ \ \frac{1}{27}\ \ \ \ or,\ \ 5^{2}\ \ =\ \ 25\ \....\ \ as\ examples.$

So, any actual numbers with powers should be reduced to their actual value, and use the identities to eliminate the negative exponents.