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CPhill
Nombre de usuario
CPhill
Puntuación
106516
Stats
Preguntas
52
Respuestas
33292
52 Questions
33362 Answers
+9
520
14
+106516
HAPPY BIRTHDAY, HECTICTAR !!!
She is one of the best "answerers" on this site [ even though she IS from Alabama....well.....everyone can't be perfect !!!! (Hehehe!!) ]
lee mas ..
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CPhill
14 oct. 2019
+41
1
1880
1
+106516
'Waiting For Moderation'
I see that this has been happening quite a bit.....
If you "right click" on the image and select "Copy Image Address".....you can open up a new browser window and paste this address into it.....hit "Enter"
lee mas ..
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CPhill
21 oct. 2018
+56
1
2155
18
+106516
4th Anniversary For Me !!!
Been on this site for 4 years now.....my thanks to Melody (and Andrew) for recommending me as a "mod".....I've learned quite a bit and "met" a lot of good people on here.....I've even managed (somehow) to help some of them....!!!
lee mas ..
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CPhill
13 mar. 2018
+22
1
1249
38
+106516
Hey, guys.....I'm going to be gone for awhile....If you can hang on....I'll answer as many questions as I know how in just a bit....
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CPhill
6 mar. 2018
+25
1060
7
+106516
I'm Out Until This Afternoon...You Kids Have Fun !!!
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CPhill
13 feb. 2018
+41
1
2286
8
+106516
HAPPY BIRTHDAY, HECTICTAR !!!!
You are a real asset to the forum.....!!!!!
Here's your cake.....enjoy.....!!!!
lee mas ..
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CPhill
14 oct. 2017
+25
1
1998
2
+106516
Going To The Picnic
This one isn't too difficult....give it a try.....!!!!!
lee mas ..
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CPhill
25 sept. 2017
+22
2
2954
7
+106516
How Much Do They Weigh ????
This is another problem from American puzzle-maker Sam Lloyd.....
Five little girls taken in couples weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116
lee mas ..
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CPhill
23 sept. 2017
+22
1
1380
5
+106516
The Width Of The River.....
Here is a problem I've always liked...it was devised by one of the great American "puzzleists" of all time, Sam Lloyd
It has been posted on the forum before, but not in a while.....maybe we have some new people who would like
lee mas ..
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CPhill
20 sept. 2017
+28
1
1853
10
+106516
ReCaptcha Verification
I don't know if others are having this problem or not......when using Opera or Firefox I had to do the ReCaptcha thing when submitting answers.......however.......using Google Chrome seems to solve this problem - for me anyway....
Comments???
lee mas ..
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CPhill
19 mar. 2017
+33
1
1948
6
+106516
Happy Pi Day..(3.14)....Everybody.....!!!!!
lee mas ..
hectictar
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CPhill
14 mar. 2017
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#1
+106516
+8
The word "of" in these types of problems usually means "multiply."
Example.......one-eighth of 40 means (1/8) * 40
Does this help??
CPhill
18 mar. 2014
#2
+106516
+5
Melody is exactly correct.......we would need more info to obtain a
specific
answer......
However........I'm going to change your question slightly to this one:
What
could be
the length of side c of a right triangle when a and b equal 3 and 4??
Well, by something known as the "triangle inequality," the remaining side is either greater than 1, or less than 7.
Then....either 4 is the longest side (the hypotenuse), or it isn't.
Let's suppose that it is......then, by the Pythagorean Theorem, the remaining side is...... SQRT(4^2 - 3^2) = SQRT(16 - 9) = SQRT(7) ........ which is greater than 1
So, if we're not too picky about whether a side is an integer or not, this could be
one
solution.
Now, let's suppose that 4 isn't the longest side. Then, again by the "P" Theorem, the remaining side is..... SQRT(3^2 + 4^2) = SQRT(9 + 16) = SQRT(25) = 5.........which is less than 7
So, we have two possible answers.......one right triangle has sides of SQRT(7), 3 and 4, where 4 is the hypotenuse. And the other triangle has sides of 3, 4 and 5, where 5 is the hypotenuse!!
I'm not sure which applies to your particular situation, but exploring possibilities is sometimes more interesting than figuring out "exact" things....
Hope this helps..
CPhill
17 mar. 2014
#2
+106516
+5
Let me expand on what Melody said.......what if we only know the sides of a triangle but we don't know any of the angles??
Well, we could use the law of cosines to find any of the included angles and then use the formula that Melody gave - a two step process.
Here's another way.....it's known as Heron's (or Hero's) formula
First....add all the sides of the triangle and divide by 2.....this is known as the semi-perimeter (S)
Then, multiply S*(S-A)*(S-B)*(S-C) where A, B, C are the side lengths of the triangle
Now take the square root of the product you just obtained, and that's the area.
The beauty of this "formula" is that there's really no trig involved at all!!
Your choice, though!!
CPhill
17 mar. 2014
#4
+106516
+5
Thanks, Melody and Alan........Both of your methods are certainly more "mathematical" than mine!!!
It's always good to learn a new method of doing things!!
CPhill
17 mar. 2014
#3
+106516
+5
Thanks, Melody.....I missed the "2 servings" part!!!
CPhill
17 mar. 2014
#1
+106516
+5
sqrt(3-x)+5=9
OK, let's first subtract 5 from both sides
This gives SQRT(3-x) = 4
Now, to get rid of the root, square both sides
We have 3-x = 16
Subtract 3 from both sides
This gives -x = 13
Multiply through by -1
We get x = -13
You should check for yourself that this is correct by putting the answer back into the original problem to make sure it makes sense.
I'll leave that for you........
CPhill
17 mar. 2014
#1
+106516
+5
Hint.....multiply both sides by 2 to isolate x
That should do it.......
CPhill
17 mar. 2014
#1
+106516
+5
Arithmetic
Consider .... 15 + 14N
If N = 1, we have 29 for the first term
If N = 2, we have 43 for the second term
etc.
CPhill
17 mar. 2014
#1
+106516
+5
110 * (2/55)
You can take it from there.....
CPhill
17 mar. 2014
#1
+106516
+5
First, convert both mixed numbers to improper fractions.
Keep the first fraction as it is, change the division to multiplication and "flip" the second fraction.........("keep"......"change"......."flip")
Now, just multiply the fractions together....(reduce if possible)
CPhill
17 mar. 2014
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