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How much does it cost to play the game? And other factors come in weather or not I have enough money. But that's irrelevant.
Although I'm not sure how this is a paradox. So far, it seems like a regular problem. I give you $2.00 and we start to play the game. Am I just being a total idiot here? Please explain the paradox part.
One morning, each member of Angela's family drank an 8 ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
Show your work please!
How did you make the rule?
I was hoping it was going to be 1234. But, I guess not!
Another problem:
Assume that
are real numbers such thatFind the value of
.My fingers are dead!!
Thanks!
Yes, I agree with 1999 for (1)
Odd.
The answer choices for (2) are (a) 30. (b) 90. (c) 450. (d) 600 and (e) 900. I think that the question for (2) also meant that Train A and B's destination are the same station that is 4 000 miles away.
Missed me? (Don't answer that)
(1) If x and y are two different numbers selected from the integers from 500 to 1000, inclusive, what is the greatest value for
(2) Train A and Train B depart at the same time from the same station for a destination 4,000 miles away. If Train A travels at a constant rate of 170 mph and Train B travels at a constant rate of 200 mph, how far will the two trains be from each other, in miles, when Train B reaches its destination?
Imagine triangle ABC. Draw an altitude from A to BC, and another one from C to BA. We'll call the place where A meets BC D and C meets AB E.
For (a)
Sue:
Isolating the bases and the altitudes gives
andMultiply those 2 equations together and do a couple of operations to get
and then divide by 2!(b) Just repeat the process used on (a)
How many liters of water must be added to 7 liters of 20% acid solution to obtain a 10% acid solution?
I have this equation where a is the liters of water, but I got a negative number.
a+20%*7=10%*(a+7)
ok thanks.
I have a question: Does this forum use Asymptote too? Asymptote is a thing where you can draw manual diagrams! It's amazing!
For Q1: Since M is the midpoint of AD, we can draw the median for trapezoid ABCD that meets CD at N. We can also draw a height from A to DC, let that be Y and the intersection of MN and AY be X. Let AB=a and CD=b
EDIT: Now with Latex!
Go to this website: http://www.mathisfunforum.com/viewtopic.php?id=22435
It has pretty much everything to the problem!
I don't know if you already thought of this, but can we also extend BD and AC to meet at point Q above AB, and then somehow use similar triangles to prove that what needs to be proven?
Don't mind Q1, I just finished solving. I got
is equal to .Thanks Bob. I realized that you could also do that by completing the square, then subtracting the same amount that you added
I have a question on this:
How can this be factored by grouping?
You get the idea... Need help with this, need help with that... blah blah bleh!
Triangle ABC is inscribed in equilateral triangle PQR, as shown. If PC = 3, BP = CQ = 2, and
, then compute AQ.In a triangle ABC, take point D on
such that DB = 14, DA = 13, DC = 4, and the circumcircles of triangles ADB and ADC have the same radius. Find the area of triangle ABC.Thanks, that is correct!
Referring to phrontister's diagram, couldn't you just instead prove that
is isosceles with ?I figured out that 9=10-1, so replacing 9 with that gives us ABCD0-ABCD=DCBA. Then use logic
This problem though... Seems impossible
I need some vital hints with these problems. I already spent like 45 minutes figuring this one out. So yeah, I'm tired
has side lengths , , and . Let be the intersection of the angle bisector of with side , and let be the foot of the perpendicular from to side . Compute the length of .Interesting...
Thanks Bobbym! You probably spent a lot of time working on these questions! Now we need Q3