Well, not specific. But after 7 hour the bus will be 210mi away
Answer 1:
It is given that the positive 2 digit number is 'x' with tens digit 't' and units digit 'u'.
So the two digit number x is expressed as,
The two digit number 'y' is obtained by reversing the digits of x.
So, 
Now, the value of x-y is expressed as:
So, 
is equivalent to (x-y).
Answer 2:
It is given that the sum of infinite geometric series with first term 'a' and common ratio r<1 = 
Since, the sum of the given infinite geometric series = 200
Therefore,
Since, r=0.15 (given)
a=170
The nth term of geometric series is given by 
.
So, second term of the series = 
= ar
Second term = 
= 25.5
So, the second term of the geometric series is 25.5
Step-by-step explanation:
If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
