The sine of 330º angle is the same as the sine of 30º. true false
1 answer:
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Answer:
- 0 ≤ t ≤ 25
- 16.348 seconds
- 310.0 meters
Step-by-step explanation:
a) Since these are production vehicles, we don't expect their top speed to be more than about 70 m/s, so the distance functions probably lose their validity after t = 25. Of course, t < 0 has no meaning in this case, so the suitable domain is about ...
0 ≤ t ≤ 25
Note that the domain for the second car would be 3 ≤ t ≤ 25.
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b) The graph of this system shows the cars will both have driven the same distance after 16.348 seconds.
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c) At that time, the cars will have driven 310.0 meters.
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<em>Non-graphical solution</em>
If you like, you can solve the equation for t:
d1 = d2
1.16t^2 = 1.74(t -3)^2
0 = 0.58t^2 -10.44t +15.66
t = (10.44 +√(10.44^2 -4(0.58)(15.66)))/(2(0.58)) = (10.44+8.524)/1.16
t = 16.348 . . . . time in seconds the cars are at the same distance
That distance is found using either equation for distance:
1.16t^2 = 1.16(16.348^2) = 310.036 . . . meters
This question is incomplete, the complete question is;
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.
In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?
Answer:
the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Step-by-step explanation:
Given the data in the question;
Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1
this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.
So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'
Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;
= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]
= [( n + 4 )!] / [ 5!( n-1 )! ]
= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120