simplify (2x^4y^0 / z^2) -5?

Question

I need to simplify this expression

Answer 1

y^0 is just 1 so it will disappear.
Remember that a power to a power is multiplication, so you multiply the exponents.
Also, you do each part separately, so the 2 is taken to the -5, the x^4, the y^0, and the z^2.
It becomes (2^-5 x^-20)/(z^-10).
Now, negative exponents are not all the way simplified, so to get rid of the negative, put the term on the other side of the fraction. the z goes to the top and the others go to the bottom
FINAL ANSWER (z^10)/(32x^20)

Answer 2

y^0 is 1 (unless y=0, in which case it's meaningless anyway). That yields 2x^4 / z^2 - 5. There isn't a significant way to further simplify this, unless you prefer to represent it as 2x^4 * z^(-2) - 5.

Answer 3

Anything to the zeroth power is 1, so y^0 drops out.

Now you have:
(2x^4/z^2 ) -5

It's unclear if you intended to raise this all to the -5 power, or not, but assuming it was, you have:
(2x^4/z^2 )^-5

Start by putting z in the numerator, but making it z^-2:
(2x^4z^-2)^-5

Distribute the exponent through:
2^(-5) * (x^4)^(-5) * (z^(-2))^(-5)

Using (a^b)^c = a^(bc)
32^-1 * x^-20 * z^10

Now swap the negative exponents to the other side:
.. z^10
----------
32 x^20

There's your final answer

Answer 4

simplify (2x^4y^0 / z^2) -5
Since y^0=1, we get (2x^4/z^2)-5, if you prefer:
(2x^4-5z^2)/z^2

Answer 5

ok i believe you meant (2x^4y^0 / z^2) ^-5
y^0 = 1 so we can just cancel that out
so you're left with
(2x^4/ z^2) ^-5
and "distributing" the -5 power
(2^-5)(x^-20)/(z^-10)
and moving negatives to the other side of the division sign
z^10/((2^5)(x^20))
(z^10)/((32)(x^20))

Answer 6

First, distribute the 2a and the 4. 2a^2 - 10a + 4a - 20 Now, upload like words. 2a^2 - 6a - 20 when you consider that 2 divides calmly into each and each and every of the numbers, component it out. 2(a^2 - 3a - 10) Now, component the equation contained in the parentheses 2 (a - 5)(a + 2) :)

Answer 7

(2x^4y^0/z^2) -5 = (2x^4 * 1 /z^2) -5
= (2x^4z^-2) - 5

Answer 8

(2x⁴y⁰ / z⁲)⁻⁵
y⁰ = 1
(2x⁴/ z⁲)⁻⁵
(2x⁴)⁻⁵ = 1/(32x⁲⁰)
(1/z⁲)⁻⁵ = 1/(1/z^10) = z^10
(2x⁴y⁰ / z⁲)⁻⁵ = (z^10)/(32x⁲⁰)

Answer 9

Ask yourself, what is y^0?