Answer:
The domain and the range of the function are, respectively:
![Dom\{f\} = [0\,m,5\,m]](https://tex.z-dn.net/?f=Dom%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%2C5%5C%2Cm%5D)
![Ran\{f\} = [0\,m^{2}, 10\,m^{2}]](https://tex.z-dn.net/?f=Ran%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%5E%7B2%7D%2C%2010%5C%2Cm%5E%7B2%7D%5D)
Step-by-step explanation:
Jina represented a function by a graphic approach, where the length, measured in meters, is the domain of the function, whereas the area, measured in square meters, is its range.
Answer:
3 is 15 degrees and 1 is 67 degrees
(2x+23)+(9x-5)=95
11x+18=95
x= 95-18/11
x=7
therefore substituting value of x in the given equations of angles:
angle ABD = 2(7)+23=37°
angle DBC = 9(7)-5= 58° Answer
to verify = 37°+58°= 95° ( which was given)
1 chip is $10
So, for 79 chips,
= 79 x $10
= $790
Answer:
16 m × 11 m
Step-by-step explanation:
The dimension of the gym is 20 m x 15 m. An outbound of 2m width is to be cut out from the gym to form the basketball court.
The original length of gym = 20 m and original width of gym = 15 m
2 m would be cut at both sides of the gym length for the outbound. Also 2 m would be cut at both sides of the gym width for the outbound. Therefore:
Length of basketball court = 20 m - (2 * 2m) = 20 m - 4 m = 16 m
Width of basketball court = 15 m - (2 * 2m) = 15 m - 4 m = 11 m
Therefore the dimensions of the basketball court is:
16 m × 11 m
