Answer
school building, so the fourth side does not need Fencing. As shown below, one of the sides has length J.‘ (in meters}. Side along school building E (a) Find a function that gives the area A (I) of the playground {in square meters) in
terms or'x. 2 24(15): 320; - 2.x (b) What side length I gives the maximum area that the playground can have? Side length x : [1] meters (c) What is the maximum area that the playground can have? Maximum area: I: square meters
Step-by-step explanation:
Answer:
y≥1
Step-by-step explanation:
hello :
y-1≥0
y-1+1≥0+1
y≥1
Explanation
This exercise is about evaluating a function at a particular argument. To do that, we replace the variable with the argument in the formula of the function, and simplify.
Let's do that:
![\begin{gathered} f(19)=\frac{3}{19+2}-\sqrt[]{19-3}, \\ \\ f(19)=\frac{3}{21}-\sqrt[]{16}, \\ \\ f(19)=\frac{1}{7}-4, \\ \\ f(19)=\frac{1-28}{7}, \\ \\ f(19)=-\frac{27}{7}\text{.} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20f%2819%29%3D%5Cfrac%7B3%7D%7B19%2B2%7D-%5Csqrt%5B%5D%7B19-3%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B3%7D%7B21%7D-%5Csqrt%5B%5D%7B16%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B1%7D%7B7%7D-4%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B1-28%7D%7B7%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D-%5Cfrac%7B27%7D%7B7%7D%5Ctext%7B.%7D%20%5Cend%7Bgathered%7D)
Answer
Answer: first, Nolan must substitute 23 for J To simplify, Nolan must subtract 16 form 23. 23 is a solution of the equation.
Step-by-step explanation:
Answer:
17/3
Step-by-step explanation:
to find out the width, we need to divide the area by the length.
68/12=5.666666667
fraction form: 17/3 (5 2/3)
