The image is blurry
Step-by-step explanation:
Answer:
It is 5/5 or 1/1 as it goes up 5 and to the left 5!!!
Answer:
x+10 x ≠ 60
Step-by-step explanation:
(x^2−50x−600)
----------------------
(x−60)
Factor the numerator
(x-60)(x+10)
-------------------
(x-60)
Cancel the like terms
x+10
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 length and width are 50 and 30 meters (or 30 and 50 meters).
Step-by-step explanation:
To solve this question, we can represent variables for the length and width in two equations.

To solve for one of the variables, you'll have to substitute one of the variables, so solve for one of them:


Now, we have a standard quadratic equation that we can factor. When factoring, you'll get this:

This tells us that the width could be either 50 or 30.
Substitute 50 into one of the equations to find the length:
2 (l) + 100 = 160
l = 30.
The length and width are 50 and 30 meters (or 30 and 50 meters).