Evaluate
\[\frac{ 1 }{ 2 }a ^{-4}b ^{2} \] b=-2 and b= 4
you have b=-2 and b=4. Do you mean a= -2 ?
Oh yes sorry xD
can you erase the "a" and put in (-2) in its place and erase b and put 4 in its place ? what do you get ?
remember that the (-2) goes with a (the whole thing is raised to the -4 power
Wait sorry I messed up\[\frac{ 1 }{ 2 }-2^{-4}4^{2}\]
Idk if I'm doing this correctly.
close, but you should put in the parens around the -2 \[ \frac{1}{2} (-2)^{-4} 4^2 \]
Now, this is a tricky exercise, because we could evaluate it directly if we had a calculator, but I suppose your teacher doesn't want that, so to do it we'll have to simplify the expression in a more simple form: \[\frac{ 1 }{ 2 }a ^{-4}b ^{2}\] \[\frac{ 1 }{ 2a ^{4} }b ^{2}\] \[\frac{ b ^{2} }{ 2a ^{4} }\] and for a=-2 and b=4 \[\frac{ 4^{2} }{ 2(-2^{4}) }=\frac{ 16 }{ -32 }=-\frac{ 1 }{ 2 }\]
There is no -1/2 in the answer options @Owlcoffee
There's only: 2 1/2 3 4
4^2 means 4*4 the exponent -4 means " flip " and change the -4 to +4 in other words \[ (-2)^{-4} = \frac{1}{(-2)^4} \]
you could try to figure it out..
what is 4*4 ?
and what is -2* -2 * -2 * -2 ?
so the answer is 1/2
4 x 4 = 16 -2 x -2 x -2 x -2 = 16
yes, so now you have \[ \frac{1}{2}\frac{16}{16} \]
and 16 divided by itself is 1
And then it'll equal to what @Owlcoffee said.
Cuz you multiply.
1 x 1/2
Here are the steps \[ \frac{1}{2} (-2)^{-4} 4^2 \\ \frac{1}{2} \frac{4\cdot 4}{(-2)^4} \\ \frac{1}{2} \frac{4\cdot 4}{(-2 \cdot -2 \cdot -2 \cdot -2)} \\ \frac{1}{2} \frac{16}{16} \\ \frac{1}{2} \]
the tricky part is knowing that the -2 goes "inside" the exponent. in other words (-2)^4 is different from - 2^4 (that first is positive, the second is negative) also, remember that a negative exponent means "flip"
Oh now I understand what you did.
I assume you know that they "invented" 2^5 as short cut to write 2*2*2*2*2 (very helpful for numbers like 10^23. ) Exponents are just a short cut to writing out ill those multiplies.
Oh yes, I know cx I just didn't understand how you got the exponent from negative to positive but now I understand cx
Do you mind helping me with one more?
@phi
Please make a new post.
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