-28xy/4
-7xy
Therefore, the first answer is correct.
Hope this helps!
Answer:
The equation 'log Subscript 3 Baseline (negative 2 x minus 3) = 2' i.e.
has x = –6 as the solution.
Step-by-step explanation:
<u>Checking the equation</u>
log Subscript 3 Baseline (negative 2 x minus 3) = 2
Writing in algebraic expression

Use the logarithmic definition








Therefore, the equation 'log Subscript 3 Baseline (negative 2 x minus 3) = 2' i.e.
has x = –6 as the solution.
Answer: A: 3x^2y^(3/2)
Step-by-step explanation:
This can be written as
(81*x^8*y^6)^(1/4)
Then multiply each exponent by (1/4):
81^(1/4)*x^(8(1/4))y^6(1/4))
81^(1/4) = 3
x^(8(1/4)) = x^2
y^6(1/4)) = y^(3/2)
The result: 3x^2y^(3/2)
7^-2= 1/7^2
= 1/14
= 0.071429