Answer:
d, e
Step-by-step explanation:
The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
a^-b = 1/a^b
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In this case, it means the product is ...
(6^1)(6^0)(6^-3) = 6^(1+0-3) = 6^-2 = 1/6^2 = 1/36
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The 6 without an exponent is equivalent to 6^1, an exponent of 1.
The sum of the exponents is -2.
Add the exponents to simplify the expression.
The value of the expression is 1/36.
An equivalent is any expression that results in 6^-2. One such is (6^5)(6^-7).
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Only the last two choices, d and e, apply.
Answer:
Infinite number of solutions.
Step-by-step explanation:
Are the equations written correctly? They are identical. Since they overlap, every point is a solution for as long as the lines reach. Since no one is stopping them, the answers are infinite.
Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.
To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.
Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.
Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.
Therefore:
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 8} \\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%208%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
The nth root of a can contain negative number only if n is an odd number.
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 2 \times - 2 \times - 2} \\ f(4) = 3 \end{cases}} \\ \large{ \begin{cases} f( - 8 ) = - 2\\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%202%20%5Ctimes%20-%20%202%20%5Ctimes%20%20%20-%202%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D%20%5C%5C%20%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20-%202%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
Answer
Answer:
The towel bar should be placed at a distance of
from each edge of the door.
Step-by-step explanation:
Given:
Length of the towel bar = 
Now given length is in mixed fraction we will convert in fraction.
To Convert mixed fraction into fraction Multiply the whole number part by the fraction's denominator, then Add that to the numerator, then write the result on top of the denominator.
can be Rewritten as 
Length of the towel bar = 
Length of the door = 
can be Rewritten as 
Length of the door = 
We need to find the distance bar should be place at from each edge of the door.
Solution:
Let the distance of bar from each edge of the door be 'x'.
So as we placed the towel bar in the center of the door it divides into two i.e. '2x'
Now we can say that;

Now we will take LCM to make the denominators common we get;

Now denominators are common so we will solve the numerators.

Or 
Hence The towel bar should be placed at a distance of
from each edge of the door.