Its true because of the rule (cpct)corresponding parts of congruence triangle.
F(3) = (1/6)3 = 1/2
The answer is 1/2 or 0.5
The trip there is 58.5$ and the trip back is 63$. 63-58.5= 4.50. Which makes the answer B
<u>Let's take this problem step-by-step</u>:
<u>Let's first set up some variables</u>:
- c: # of children
- a: # of adults
<u>Let's examine the information given:</u>
- Elevator can hold a maximum of 1500 pounds
⇒ average child is 75 pounds
⇒ average adult is 150 pounds
⇒<em> therefore</em>: 
- Elevator can fit no more than 14 people
⇒ <em>therefore</em>: 
<u>Let's graph the equations</u>:
⇒ look at the image attached
<u>The point at which the two graphs intersect:</u>
⇒ <em>is the solution that represents the amount of children and adults and </em>
<em> their combine weight</em>
<em />
<u><em>With the horizontal axis being the # of children and vertical axis being the # of adults</em></u><em>:</em>
<em> ⇒ the </em><em>solution is 8 children and 6 adults</em>
<em></em>
<u>Answer: 8 children and 6 adults</u>
<u></u>
Hope that helped!
<em />
Do you mean no repeating numbers within the two sets? Because if so, I don't think it's possible.
I started by trying to figure out what the numbers in the tenths place should be. I used subtraction: 71 - ___ = ____. If you try it out, you can't subtract anything with a 6, 7, 8, or 9 in the tenths place because it will leave you with 11 (a repeating digit number), 10 (has a 0), or less (1-digit numbers). Also, a 3 cannot go into the tenths place because when you do 71 - 3_ , your answer will always begin with a 3 (problem because the 3 repeats), or it will contain a 0.
Therefore, the numbers left for the tenths place are: 1, 2, 4, and 5. 1 and 5 pair up, leaving 4 and 2.
71 - 5_ = 1_ and 71 - 4_ = 2_.
Then, I tried to figure out what numbers go in the ones place. That lead me to realize they act in pairs. The pairs possible in the ones place are 2 and 9, 3 and 8, 4 and 7, 5 and 6. These numbers always go together to result with the final "1" in the "71". Using this information, I looked at the numbers I already used: 1, 2, 4, and 5. Now, looking at the pairs, I eliminated the pairs containing a number already used. This leaves me with only one pair: 3 and 8. Obviously, you need two more pairs to solve the problem, which leads me to my point of saying: This problem is impossible to solve.
I really hope someone can prove me wrong! But this is the solution I have reached for now. :)
