We need to compute for the length of each trophy given that one trophy is 5 inches wider than the other.
Let x be the length of the first trophy.
Let x+5 be the length of the second trophy
The solution is shown below:
x + (x + 5) = 17
x + x + 5 =17
2x =17 -5
2x = 12
x = 12/2
x=6 (for the first trophy)
x + 5 = 6+5 = 11 (for the second trophy)
The first trophy is 6 inches wide while the second trophy is 11 inches wide.
H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t
A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).
B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.
What is the half of the number
Answer:
look up math way put in your decimal and ask it to convert into a fraction
Step-by-step explanation: