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CPhill
Nom d'utilisateur
CPhill
But
130070
Membership
Stats
Questions
56
Réponses
43251
56 Questions
43668 Answers
+16
29
7
+130070
Blocked Posts
I just wrote to the site manager to let him know that this is happening....I'll try to "unblock" as many posts as possible ( but realize that I can't stay on here all day )
CPhi
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CPhill
7 août 2023
+18
1
1854
32
+130070
HAPPY BIRTHDAY , MELODY !!!
I baked you a cake.....be sure to share !!!!!
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CPhill
16 avr. 2022
+25
3624
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HAPPY BIRTHDAY, MELODY !!!!
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CPhill
16 avr. 2021
+57
4488
4
+130070
Gotta take a break folks !!!!
I'll come back a little later and try to answer as many questions as I can [ or as many as I know how to do !!! ]
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CPhill
26 mars 2020
+46
1
5458
14
+130070
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!!) ]
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CPhill
14 oct. 2019
+54
2
4313
1
+130070
'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"
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CPhill
21 oct. 2018
+84
2
4497
18
+130070
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....!!!
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CPhill
13 mars 2018
+26
1
2937
38
+130070
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 mars 2018
+32
1
2435
7
+130070
I'm Out Until This Afternoon...You Kids Have Fun !!!
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CPhill
13 févr. 2018
+48
2
4181
8
+130070
HAPPY BIRTHDAY, HECTICTAR !!!!
You are a real asset to the forum.....!!!!!
Here's your cake.....enjoy.....!!!!
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CPhill
14 oct. 2017
+31
3
3978
2
+130070
Going To The Picnic
This one isn't too difficult....give it a try.....!!!!!
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CPhill
25 sept. 2017
+29
2
6914
7
+130070
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
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CPhill
23 sept. 2017
+26
2
2921
5
+130070
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
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CPhill
20 sept. 2017
+31
2
3232
10
+130070
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???
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CPhill
19 mars 2017
+35
1
3786
6
+130070
Happy Pi Day..(3.14)....Everybody.....!!!!!
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hectictar
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CPhill
14 mars 2017
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#1
+130070
+2
This represents 52 ! where the "!" is known as "factorial"
In this case.....it's a REALLY BIG number !!
52!
See....I told you so !!!!
CPhill
27 mars 2014
#2
+130070
+2
Isn't your equation
y = 3.75 + .75(x)
where 3.75 is the "fixed cost" part and .75(x) the "variable cost" part ???
If so, we have
14.50 = 3.75 + .75(x)
Subtracting 3.75 from both sides, we have
11.25 = .75(x)
Dividing both sides by .75, we have
x =
11.25/.75
(This usually represents some variable quantity like hours worked, etc.)
CPhill
27 mars 2014
#1
+130070
+2
The sum of the first N positive integers is given by N * (N+1) / 2
So we have......
401* 402 / 2
Notice that we could also see a pattern here.....
400 + 1 = 401
399 + 2 = 401
398 + 3 = 401
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201 + 200 = 401
So, we have 200 pairs that add to 401, or
200(401) plus the lone remaining "unpaired" 401 = 80601
That's it !!!
CPhill
27 mars 2014
#1
+130070
+2
At the end of the first year, you get 100% of the 3300 plus 5% more.
In decimal terms, this equates to 1.00 + .05 = 1.05.
So, to find the amount of the raise after the first year, we just multiply 3300 * (1.05)
To find the amount after two years, we take what we had at the end of the first year (3300 * (1.05)) and multiply THAT by 1.05.
So we have........(3300 * 1.05) * (1.05 or 3300 * (1.05) ^2
Then, the "formula" to find the amount after N years is just........... 3300*(1.05)^N
I'll let you finish this one.....(remember to do the exponent part first and then multiply by 3300, unless you're putting the whole thing in a calculator to evaluate it all at once)
CPhill
27 mars 2014
#3
+130070
+2
This isn't the most "mathematical" explanation.....but it might help anyway.
Let us suppose that something originally costs $100.
Now, let's suppose that we take 10% off. Well, looking at it another way, it must now just be "worth" 90% of what its original price was, i.e., $90.
Now, let's suppose that we take 15% off of that. Well, then the "new" price must just be 85% of the $90 (since we took 15% off), i.e., $76.50.
Well, the "final" price is just 76.5% of the original $100 price........thus we "took off" $23.50.....or 23.5% of 100 !!!
In terms of "math-speak," we could express this in the following way........100 - (100*.90*.85)......100 - (76.50)........Original Price - (Final Price) = "Discount"
This is simplistic, but I hope it helps you see it.......
CPhill
27 mars 2014
#1
+130070
+2
We could use some fancy math here, but let's not.
Since a boy 6 feet tall casts a shadow 4 times his height (24 ft.), the flagpole must do the same.
Then the flagpole's shadow must be 4 times its height.
Then dividing 200 by 4, the height of the flagpole must be 50 ft.
And that it. No fuss, no muss.
CPhill
27 mars 2014
#1
+130070
+2
We could use some fancy math here, but let's not.
Since a boy 6 feet tall casts a shadow 4 times his height (24 ft.), the flagpole must do the same.
Then the flagpole's shadow must be 4 times its height.
Then dividing 200 by 4, the height of the flagpole must be 50 ft.
And that it. No fuss, no muss.
CPhill
27 mars 2014
#1
+130070
+2
X = 36 degrees
Call the apex point of the triangle A
Call the angle bisector AD, where D is the point where the bisector intersects the base of the triangle.
Let the vertexes counterclockwise from A be labeled B and C.
Then, in triangle ADC, AD = DC. And in any triangle having equal sides, the angles opposite those sides are equal, too. Therefore, < DCA (angle "X") = < DAC (one of the angles formed by the bisector).
And in triangle ABD, AB = AD, so the measure of angles ABD and ADB are equal.
And by the exterior angle theorem, angle ADB = DAC + DCA. Therefore, ABD = BAD + DCA.
And, because of bisection, angle BAD = DAC. And angle BAC equals their sum.
Therefore, ABD + BAC + DCA = 180.
Therefore, by substitution, BAD + DCA + BAD + DAC + DCA = 180 or, simplifying, 2BAD + 2DCA + DAC = 180.
But DCA = DAC., And DAC = BAD.
Therefore, DCA = DAC = BAC = BAD.
And substituting again......2DCA + 2DCA + DCA = 180
Therefore, 5DCA=180
Therefore DCA = 36 = X
Let's check that this is true. In triangle ABD....ABD = 72 angle ADB =72 and BAD = 36....these sum to 180
And, in triangle ADC, angle ADC is supplemental to angle ADB (i.e., 108) and DAC = 36 and DCA = 36...so these sum to 180, as well.
CPhill
27 mars 2014
#1
+130070
+2
Have you tried the calculator on this site??
Hit the "2nd" button on the calculator and find the "asin" function.
You also have the option to express this in degrees or radians.
If you still have trouble - let us know.
CPhill
27 mars 2014
#1
+130070
+2
what is 526923076.923 in scientific notation?
5.26923076923 × 10^8
CPhill
26 mars 2014
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