Answer:
30 s
Step-by-step explanation:
When the ball hits the ground h=0. To find the time t when this happens we must solve the equation h=0.
●h= 0
● -12t^2+360t =0
● t(-12t +360) = 0
● t = 0 or -12t +360 =0
● t=0 or -12t = -360
● t=0 or 12t =360
● t=0 or t=360/12
● t=0 or t= 30
The equation has two solutions.
The ball was fired with an initial speed of 800 feet per second so it cannot hit the ground at t=0.
So the ball hits the ground after 30 s.
Answer:
see explanation
Step-by-step explanation:
We require 2 equations with the recurring part after the decimal point.
Let x = 0.555..... → (1)
Multiply both sides by 10, then
10x = 5.555..... → (2)
Subtract (1) from (2) thus eliminating the recurring decimal
(10x - x) = (5.5555 - 0.5555 ), that is
9x = 5 ( divide both sides by 9 )
x = 
717/3= 239
Divide 717 by 3 to get the price of one night.
Part A
Answer: The common ratio is -2
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Explanation:
To get the common ratio r, we divide any term by the previous one
One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2
Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time
Side Note: each term is multiplied by -2 to get the next term
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Part B
Answer:
The rule for the sequence is
a(n) = (-2)^(n-1)
where n starts at n = 1
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Explanation:
Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio
The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)
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Part C
Answer: The next three terms are 16, -32, 64
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Explanation:
We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms
-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64
showing that the next three terms are 16, -32, and 64
An alternative is to use the formula found in part B
Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 .... which matches with what we got earlier
Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 .... which matches with what we got earlier
Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 .... which matches with what we got earlier
while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)
