Question

Can someone help me with expressions with exponents pleaseeeeeeeeee?

Answer 1

You need more e's in your pleeeeeeeeeeeeeeeeeeeeeeeaseeeeeeeeeeeeeeeeeeeeeee! ;P

Answer 2

here are some!

Answer 3

pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Answer 4

For that first one:

Answer 5

Simplify the following:\[-5×4 x^{3} x^{2} \]Combine products of like terms.
\[4 x^3 (-5) x^{2} = 4 x^{(3+2)} (-5): \]\[-5×4 x^{(3+2)}\]Evaluate 3+2.
\[3+2 = 5: \]\[-5×4 x^5 \]Multiply 4 and -5 together.
\[4 (-5) = -20: \]

Answer 6

Now solve the equation.

Answer 7

thanks but I just need to simplify it & remove all negative and zero exponents, not solve the equation but thanks for the work you did :)

Answer 8

hmmm he was "simplifying" it, thus

Answer 9

Yes, that's what I was doing haha. I meant solve the equation I gave you. You should have gotten -20x^5 as your answer.

Answer 10

ohh wait I think I got it now. Thanks :) I understand that first one!

Answer 11

No problem :) and for the second one:

Answer 12

I tried the second one on my own and I got -12 x^3 y^3
is that correct or?

Answer 13

Simplify the following:
\[-3×4 x^2 y x y^2 \]Combine products of like terms.
\[4 x^2 y (-3) x y^2 = 4 x^{(2+1)} y^{(1+2)} (-3):\]\[-3×4 x^{(2+1)} y^{(1+2)}\]Evaluate 1+2.
\[1+2 = 3: \]\[-3×4 x^{(2+1)} y^3 \]Evaluate 2+1.
\[2+1 = 3: \]\[-3×4 x^3 y^3 \]Multiply 4 and -3 together.
\[4 (-3) = -12: \]

Answer 14

No, you should have gotten \[-20x^5\] Where do you think you went wrong?

Answer 15

Oh wrong one hahaha

Answer 16

Yes, you got it right :)

Answer 17

thanks :)))

Answer 18

For that third one:

Answer 19

Simplify the following:
\[ \frac{ 2x^{2} }{ x }\]For all exponents, \[\frac {a^{n}} {a^{m}} = a^{(n-m)}\]. Apply this to\[ \frac{(2 x^2)}{x}. \]Combine powers. \[\frac {2 x^2}{x} = 2 x^{2-1}:\]\[2 x^{2-1} \]Evaluate 2-1.
\[2-1 = 1: \]

Answer 20

And what will your answer be?

Answer 21

2x^1 ?

Answer 22

Can \[x^1\] be simplified?

Answer 23

just as 2x?

Answer 24

Yep, nice job.

Answer 25

thanksss

Answer 26

No problem. For number four:

Answer 27

Simplify the following:
\[\frac{2 x^2}{8 x^2}\]Distribute exponents over products in \[(2 x)^2. \]
Multiply each exponent in 2 x by 2:
\[\frac{2^2 x^2}{8 x^2}\]
Cancel common terms in the numerator and denominator of \[\frac{2^2 x^2}{8 x^2}.\]
\[\frac{2^2 x^2}{8 x^2} = \frac{x^2}{x^2}×\frac{2^2}{8}= \frac{2^2}{8}:\]2^2/8
Evaluate 2^2.
\[2^2 = 4: \]
\[\frac{4}{8}\]Reduce 4/8 to lowest terms. Start by finding the GCD of 4 and 8.
The gcd of 4 and 8 is 4, so \[\frac{4}{8} = \frac{4×1}{4×2} = \frac{4}{4}×\frac{1}{2} = \frac{1}{2}:\]

Answer 28

What's your answer?

Answer 29

I don't understand why it's 2^2 x^2 instead of just 2x^2 :/

Answer 30

Because you're multiplying all of your exponents by two to keep your numerator from getting jumbled up when you simplify.

Answer 31

I'm still kind of confused, so there's no way to use 2x^2?

Answer 32

You're using 2^2 x^2 to simplify 2x^2 because it makes x dividable by 2 when you simplify 8/4. You can skip that step if you'd like and just make it 2*2/8, but that wouldn't work on every other problem you did.

Answer 33

ok i'll use what you showed me and both the x^2 cancel out right? I just got 1/2 ..

Answer 34

Mhm, that's correct.

Answer 35

thanks, but applying the exponent to both the variable and whole number doesn't work on every problem right?

Answer 36

There's not much to do in number five, just solve:\[-3^{-4}=x \]\[\frac{1}{x}\]

Answer 37

Right.

Answer 38

I'm a little confused as to where the x came from?

Answer 39

The x is to transfer \[-3^{-4}\]to the simplified fraction. I put that in there. The answer would just be \[\frac{1}{-3^{-4}}\]

Answer 40

so you just did the reciprocal since they're negative?

Answer 41

that's all you do?

Answer 42

Yep, the simplified version would be \[\frac{1}{81}\]

Answer 43

so the answer is 1/81? since you did the reciprocal do both the whole number and exponent become positive or just the exponent ?

Answer 44

When you raised -3 by -4 you automatically got 81 because \[-x^{-y}=z\]\[x^{y}-z\]and\[-x^{y}=-z\]\[x^{-y}=-z\]

Answer 45

I meant "=" not "-" on that second one.

Answer 46

ahh ok :)

Answer 47

is #16 1y^-4 ?