The cost of an adult ticket is £6 more than that of a child ticket, so will be denoted by c+6. Now, we are told that the cost of four child tickets and two adult tickets is £40.50, so we can put this in an equation and solve for c:
(c+6)+(c+6)+c+c+c+c=40.50
6c+12=40.50
6c=28.50
c=4.75
Therefore the cost of a child's ticket (c) is £4.75 and the cost of an adult ticket (c+6) is £10.75.
We need to assign a value for x to check the possible values of y.
1st inequality: y < -0.75x
X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT
X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN
X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT
2nd inequality: y < 3x -2
X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT
X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT
X = 1 ; y < 3(1) – 2 ; y <<span> 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT
The actual solution to the system lies on the 4TH QUADRANT.
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To find the time at which both balls are at the same height, set the equations equal to each other then solve for t.
h = -16t^2 + 56t
h = -16t^2 + 156t - 248
-16t^2 + 56t = -16t^2 + 156t - 248
You can cancel out the -16t^2's to get
56t = 156t - 248
=> 0 = 100t - 248
=> 248 = 100t
=> 2.48 = t
Using this time value, plug into either equation to find the height.
h = 16(2.48)^2 + 56(2.48)
Final answer:
h = 40.4736
Hope I helped :)
