“B” is the answer I think
Answer:
the answer is let a b* ban me
Step-by-step explanation:
Answer:
Step-by-step explanation:
16x^4y^-3z^4 / 36x^-2yz^0
16x^4y^-3z^4 = 16x^4z^4/y^3
36x^-2yz^0 = 36x^-2y(1) =36x^-2y = 36y/x^2
16x^4y^-3z^4 / 36x^-2yz^0
= (16x^4z^4/y^3) / (36y/x^2)
= 16x^4z^4/y^3 * x^2/36y
= (4/9)x^6z^4/y^4
or another way
fist multiply it out
f(x) = 4x^(3/5) - x^(8/5)
now differentiate knowing d/dx(x^n) = n x^(n-1)
to get
4*(3/5) x^(-2/5) - 8/5 x^(3/5)
simplify to get
12/5/x^(2/5) - 8/5 x^(3/5)
If this is what your looking for please give me brainiest, i have done this problem in the past so i know how to solve it :)
The second description is not suitable.
It should be "simplify the terms" sice you are just adding 36 and -20.
Consider the given expression,
Apply the distributive property,
Combine like terms,
Subtract 20 from both sides,
Divide both sides by 27,
Answer:
x = 10
Step-by-step explanation:
You can try the answers to see which works. (The first one does.)
Or, you can solve for the variable:
Divide by 75
... (1/5)^(x/5) = 3/75 = 1/25
Recognize that 25 = 5^2, so ...
... (1/5)^(x/5) = (1/5)^2
Equating exponents, you have
... x/5 = 2
... x = 10 . . . . . multiply by 5
_____
You can also start by taking logarithms:
... log(75) +(x/5)log(1/5) = log(3)
... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log
... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression
... x = 10 . . . . . multiply by 5
_____
"Equating exponents" is essentially the same as taking logarithms.
