The answer would be B, because you would just multiply 116.22 by 500
Answer:
(a) yes it does
(b) the rate is constant because the rate of change (slope) is the same throughout the function.
(c) The rate, or rate of change, is -1/2.
y1 - y2/x1-x2
4 - 2/3 - 9
2/-6
-1/2
Answer:
(a) B
(b) $2
Step-by-step explanation:
(a) Let's say the cost of a ticket is t and the cost of popcorn is p. Then we can write the two equations from the table:
12t + 8p = 184
9t + 6p = 138
We need to solve this, so let's use elimination. Multiply the first equation by 3 and the second equation by 4:
3 * (12t + 8p = 184)
4 * (9t + 6p = 138)
We get:
36t + 24p = 552
36t + 24p = 552
Subtract the second from the first:
36t + 24p = 552
- 36t + 24p = 552
________________
0 = 0
Since we get down to 0 = 0, which is always true, we know that we cannot determine the cost of each ticket because there is more than one solution (infinitely many, actually). The answer is B.
(b) Our equation from this, if we still use t and p, is:
5t + 4p = 82
Now, just choose any of the two equations from above. Let's just pick 9t + 6p = 138. Now, we have the system:
5t + 4p = 82
9t + 6p = 138
To solve, let's use elimination again. Multiply the first equation by 6 and the second one by 4:
6 * (5t + 4p = 82)
4 * (9t + 6p = 138)
We get:
30t + 24p = 492
36t + 24p = 552
Subtract the second from the first:
36t + 24p = 552
- 30t + 24p = 492
________________
6t + 0p = 60
So, t = 60/6 = $10. Plug this back into any of the equations to solve for p:
5t + 4p = 82
5 * 10 + 4p = 82
50 + 4p = 82
4p = 32
p = 32/4 = $8
So the ticket costs 10 - 8 = $2 more dollars than the popcorn.
Answer:
x = -2
Step-by-step explanation:
2x + 1 + 3 = 0
2x + 4 = 0
x = -2
Answer:
5 or less
Step-by-step explanation:
The speed increased linearly with distance, but is not decreasing linearly with distance. This suggests the track has an unknown shape, so prediction of car behavior is a guess, at best.
If the car continues to decrease its speed at 3 units per unit distance, then the final 3 units of speed will decrease to 0 in one additional unit of distance. That is, the car will stop at a distance of 5 units.
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Since the car has be decreasing its speed at an increasing rate with respect to distance, very possibly the car will stop before it reaches distance unit 5.
Since we don't know the track shape, it seems possible the car may not stop until some large unknown number of distance units, say 10 or 1000.
