Answer:
Luis’s, because he flipped the inequality sign when he subtracted
Step-by-step explanation:
Given:
7.2b + 6.5 > 4.8b – 8.1.
Amelia started by subtracting 7.2b from both sides to get 6.5 > –2.4b – 8.1.
Amelia:
7.2b + 6.5 - 7.2b > 4.8b – 8.1 - 7.2b
6.5 > -2.4b - 8.1
Correct
Luis started by subtracting 4.8b from both sides to get 2.4b + 6.5 < – 8.1.
7.2b + 6.5 - 4.8b > 4.8b – 8.1 - 4.8b
2.4b + 6.5 > -8.1
Incorrect
Shauna started by subtracting 6.5 from both sides to get 7.2b > 4.8b – 14.6
7.2b + 6.5 - 6.5 > 4.8b – 8.1 - 6.5
7.2b > 4.8b - 14.6
Correct
Clarence started by adding 8.1 to both sides to get 7.2b + 14.6 > 4.8b.
7.2b + 6.5 + 8.1 > 4.8b – 8.1 + 8.1
7.2b + 14.6 > 4.8b
Correct
How come it’s a right triangle, I think you do a2+b2=C2. Therefore, you do 9.4 times 9.4 and then 15.1 times 15.1. So it will be 9.4 + b2=15.1. 88.36+ b2=228.01. Then you subtract that and it will be 139.65. Then you do square roots because you are trying to get rid of the second power. And the final answer is 11.81736011. How they told you to round it to the nearest tenth, I will be 11.8. I think this is the answer.
You didnt list any graph?? Can you add it please :)
3) 2 to 3
4) 4 to 5
5) 5 to 6
6) 5 to 7
ur welcome kid
301
We could start by finding the lowest common multiple of 2, 3, 4, 5, and 6, which is 60. Then, we can consider the next few multiples: 120, 180, 240, 300...
However, because we need a remainder of 1 when our number is divided by each of these numbers (2,3,4,5,6), we want to go one above each of these multiples. So we're talking about 61, 121, 181, 241, 301... Those are the numbers that will satisfy the "remainder of 1" part of the question.
Now, we need to find out which one satisfies the other part of the question, which just requires dividing each of these numbers by 7 to see which is divisible by 7 (in other words, which one gives us a remainder of zero when we divide by 7).
301 does it. 301/7 = 43. So 301 is a multiple of 7 and therefore will yield no remainder when divided by 7.
Hope this all makes sense.
