Without drawing the graphs, find the points of intersection of the lines: a 7x+4y=23 and 8x–10y=19
1 answer:
Answer:
(3,0.5)
Step-by-step explanation:
The given system has equations:
Let us multiply the second equation by 4 and the first equation by 10.
We add both new equations to get:
Put x=3 into the first equation and solve for y.
The point of intersection is the solution to the system of equation which is (3,0.5)
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Answer:
This is a proportional relationship, the constant of proportionality is 20m/s and it represents that the horse can run 20 meters every second.
Equation: d = 20s, where d=distance and s=number of seconds.
Step-by-step explanation:
In order to find out whether this relationship is proportional, you need to see if the rate at which the horse runs is constant (the same). If you look at the three sets of data (24, 480), (40, 800) and (60, 1200) where the pair is (seconds, meters), you can see that for any two sets of data the change in meters divided by the change in seconds is consistently 20m/s. For example:
Since the constant is 20, we know that the horse can run 20 meters every second. To find the horse's total distance, we need to multiply the rate by the number of seconds that it runs:
d = 20s
We need to find the number of integers between 100 and 500 that can be divided by 6, 8, or both. Now, to do this, we must as to how many are divisible by 6 and how many are multiples of 8.
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101