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.
Since there are 2 black suites and there are 26 (1/2) black cards in a deck, and there is 1 five per suite, there are 2 black fives. Therefore, since there are 52 cards in a deck, there is a 2/52 you draw a black 5. To find the odds against that, we get 1-2/52=52/52-2/52=50/52 due to that we want to find the probability of not drawing a black 5 so we subtract the chances of getting that from the equation
Answer: 47
Step-by-step explanation:
simply substitute the constants with 5 and 3.
(5)^2 + 9(3) - 5 = 47
1.<u>3 </u>x 1= 1.<u>3 </u>hope this helped
