Determine the ordered pair that satisfies the equation, 5x - 3y = 2.
1 answer:
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Answer:
A- Triangle 2: {8, 15, 17}
Step-by-step explanation:
You may recognize that the triple of choice A is a Pythagorean triple, so can be the side lengths of a right triangle.
Checking the numbers can be done nicely by a graphing calculator or spreadsheet to help avoid the tedium of doing it "by hand".
The basic idea is that the triangle will be a right triangle if the sum of the squares of the shorter sides is the square of the longer side.
Answer:
Step-by-step explanation:
We are given the expression:
There are no common factors between the four numbers, however the first two have a common factor and the last two do too. Therefore, we can factor by grouping.
Group the first two terms and the last two.
Find the greatest common factor (GCF) of the first group. It is 5g². Factor it out of the first group. You can do this by dividing both terms by the GCF.
- 10g³/5g²= 2g
- -25g²/5g² = -5
Repeat with the second group. The GCF is 2.
- 4g/2 = 2g
- -10/2= -5
There is another GCF: 2g-5. We can factor this out of both terms.
- 5g²(2g-5)/2g-5= 5g²
- 2(2g-5)/ 2g-5=2
This cannot be factored further, so it is the answer.
Recall your d = rt, distance = rate * time.
now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.
if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.
now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.
now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.
however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.
now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.
if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.
now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.
now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.
however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.