Indices: Simplofy (show all working )?

Question

A) (18a)^0 G)3y^2 multiply 2y^0

B) 3c multiply 4cd

C) (2 divivde5)^5

E) x^b multiply x^c

F)14x^3y^5 divide7x^2y^2

Answer 1

A) Provided a is not zero, (18a)^0 = 1. (Any non-zero quantity raised to the power 0 is 1. 0^0 is UNDEFINED.)

B) 3c*4cd = 12 c^2 d. (Just multiply and combine like terms.)

C) (2/5)^5 = 2^5 / 5^5 = 32 / 3125 = 0.01024.

(Some would argue that (2/5)^5 WAS the simplest form.)

D) There IS no D)!

E) x^b*x^c = x^(b + c). (That's just how indices work; on multiplying, you ADD the indices of a common base; on dividing, you SUBTRACT the indices.)

F) 14x^3y^5 / (7x^2y^2) = 2 x y^3. (Divide like factors, using the dividing rule for exponents described in part E.)

G) Since 2 y^0 is simply 2, assuming y is not zero (see A), this is 6 y^2. If y = 0, it's undefined.

QED

Live long and prosper.

NOTE. Despite what the next responder says, it's NOT true that "Anything to the 0 power is 1," IF that "Anything" happens to be zero.

The reason for that is that the PROOF that a^0 = 1 for non-zero ' a ' rests on the following kind of argument, which is best seen for a specific value of ' a, ' a = 2 for simplicity:

O.K., let a = 2, then 2^5 = 32, 2^3 = 8, 2^1 = 2, so a natural progression there says that we're dividing by 2^2 (= 4) each time so that 2^(-1) = 1/2, 2^(-3) = 8, etc. We see that (2^3)*[2^(-3)] = 8 x 1/8 = 2^(3 - 3) = 2^0 = 1, as it should, and everything is fine.

If however we try that same argument with 0^5 = 0, 0^3 = 0, 0^1 = 0, the next term in the analogous series is 0^(-1). But that OUGHT to be 1/0, which is UNDEFINED, as is the next term 1/0^3, etc. So, if you try to show that 0^0 must be 1 by the analogous argument to the one we used before, it doesn't work because we'd have to multiply 0^3 (say), which IS defined, by 0^(-3), which ISN'T defined.

This problem arises at the first point where our previous argument fails to work, and that point is at the moment when we try to give a meaning to 0^0. No matter how you try to get around it, you have to face the fact that 0 doesn't work like any NON-ZERO base. The quantity 0^(n^2) for real n exists and is zero, but 0^(- n^2) is UNDEFINED, and therefore their PRODUCT 0^0 is UNDEFINED.

Answer 2

Question A
1 x G x 3y² x 2 = 6 G y ²

Question B
12 c ² d

Question C
2^5 / 5^5 = 32 / 3125

Question E
x^(b + c)

Question F
2 x y ³

Answer 3

I scanned it again ^^; It's easier for me to scan.

Here:
http://img256.imageshack.us/img256/2889/math4001ln8.jpg

Notes:
-No (D)? Okay. Lol.
-Anything to the 0 power is 1
-When two exponents have the same base (refer to E) are multiplied, you add the exponents. When they are divided (like x^5/x^3) you subtract the exponents.

&& As always, IM or email or let me know if you need help or explanation. kthxbye.