Answer:
Z + 12
Step-by-step explanation:
Z-(-12)
Z + 12
Double negative turns into positive
Question:
A cinema can hold 270 people at one performance 5/9 of the seats were occupied of the occupied seats 40% we occupied by concessionary ticket holders.
What is the number of seats occupied by concessionary ticket holders?.
Answer:
60 seats
Step-by-step explanation:
Given
Number of seats = 270
Occupied Seats = 5/9
Concessionary ticket holders = 40% of occupied Seats
Required
The number of seats occupied by concessionary ticket holders
First the number of occupied seat has to be calculated.
![Occupied\ Seats = \frac{5}{9} * 270](https://tex.z-dn.net/?f=Occupied%5C%20Seats%20%3D%20%5Cfrac%7B5%7D%7B9%7D%20%2A%20270)
![Occupied\ Seats = \frac{1350}{9}](https://tex.z-dn.net/?f=Occupied%5C%20Seats%20%3D%20%5Cfrac%7B1350%7D%7B9%7D)
![Occupied\ Seats = 150](https://tex.z-dn.net/?f=Occupied%5C%20Seats%20%3D%20150)
Next is to determine the number of seats occupied by concessionary ticket holders.
![Number = 40\%\ of\ occupied\ seats](https://tex.z-dn.net/?f=Number%20%3D%2040%5C%25%5C%20of%5C%20occupied%5C%20seats)
![Number = 40\%\ of\ 150](https://tex.z-dn.net/?f=Number%20%3D%2040%5C%25%5C%20of%5C%20150)
Convert percentage to decimal
![Number = 0.4 * 150](https://tex.z-dn.net/?f=Number%20%3D%200.4%20%2A%20150)
![Number = 60](https://tex.z-dn.net/?f=Number%20%3D%2060)
<em>Hence, 60 seats were occupied by concessionary ticket holders.</em>
Answer: what the heck do you have to answer
Step-by-step explanation:
Answer:
Range of Function : { - 9, - 5, - 1, 4 }
Step-by-step explanation:
We know that y = 2x - 5 provided the domain ( x - values ) { - 2, 0, 2, 4 }. Let us substitute each element in this set of domain as x in the equation "y = 2x - 5" as to solve for the y - values, otherwise known as the range of the function.
{ - 2, 0, 2, 4 }
y = 2( - 2 ) - 5 = - 9,
y = 2( 0 ) - 5 = - 5,
y = 2( 2 ) - 5 = - 1,
y = 2( 4 ) - 5 = 4
We have the set of y - values as { - 9, - 5, - 1, 4 }. This is the range of our function.
You are given two equations, solve for one variable in one of the equations. Say you solved for x in the second equation. Then, plug in that value of x in the x of the first equation. Solve this (first) equation for y (as it should become apparent) and you'll get a number value. Plug in this numerical value of y into the y of the second equation. Solve for x in the second equation. And there you have it: (x, y)