Answer:
9/20
Step-by-step explanation:
Total trials = 80
Observed successful trials = 36
36/80
= 9/20
Answer:
- <u>0.5 s</u> (using 1 significan figure since the initial velocity is given with 1 significant figure)
Explanation:
Let's see what each term in the function v (t) = - 9.8t + 5 means:
The variable t is the time
The coffecient - 9.8 (which is negative) means that the velocity is reduced 9.8 m/s each second.
That means that the motion is uniformly decelerated with acceleration - 9.8 m/s².
The term + 5, means that the object was launched with an upward velocity of 5 m/s.
Hence the equation tells that after the object was launched upward it starts to lose speed and will reach a maximum height when the velocity is equal to zero, from which it starts falling.
So, you can calculate when the object reachs it maximum height and starts falling by making the velocity, v (t), equal to zero and solving for t. This is how you do it:
- v (t) = 0
- 0 = - 9.8t + 5
- 9.8t = 5 remember that the units of 9.8 is m/s² and of 5 is m/s
- t = 5 m/s / 9.8 m/s² = 0.51 s
So, the object starts fallin at 0.51 s. Since, the velocity is reported with 1 significant digit you answer should also show 1 significant digit, which means t = 0.5 s.
Answer:
12
Step-by-step explanation:
The first cave has 7 times more bats than the last cave. So if the 45th cave has b bats, then the first cave has 7b bats.
There are 77 bats in every row of 7 caves. So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.
Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats. Repeating this logic:
#1 = 7b
#2-#7 = 77−7b
#8 = 7b
#9-#14 = 77−7b
#15 = 7b
#16-21 = 77−7b
#22 = 7b
#23-28 = 77−7b
#29 = 7b
So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.
Now we do the same thing from the other end. If cave #45 has b bats, then caves #39-#44 have 77−b bats. And since caves #38-44 have 77 bats, then cave #38 has b bats. Therefore:
#45 = b
#39-44 = 77−b
#38 = b
#32-37 = 77−b
#31 = b
So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.
Adding that to the first 29 caves, plus x number of bats in cave #30:
308 + 7b + x + 154 + b = 462 + 8b + x
We know this equals 490.
490 = 462 + 8b + x
28 = 8b + x
x is a maximum when b is a minimum, which is b = 2.
28 = 8(2) + x
x = 12
There are at most 12 bats in the 30th cave.
Multiply all 4 by each other.
Answer:
Part A:
Two types of translation are;
1) Horizontal translation left T(0, 8),
2) Vertical translation T(16, 0)
Part B:
For the horizontal translation transformation, k = 8
For the vertical translation transformation, k = 16
Part C:
For the horizontal translation transformation, the equation is f(x + 8) = g(x)
For the vertical translation transformation, the equation is f(x) + 16 = g(x)
Step-by-step explanation:
