член: CPhill

56 Questions
43598 Answers

+13

+129899

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 прочитайте больше ..

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CPhill 7 авг. 2023 г.

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HAPPY BIRTHDAY , MELODY !!!

I baked you a cake.....be sure to share !!!!!

прочитайте больше ..

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CPhill 16 апр. 2022 г.

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HAPPY BIRTHDAY, MELODY !!!!

прочитайте больше ..

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CPhill 16 апр. 2021 г.

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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 !!! ]

прочитайте больше ..

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CPhill 26 мар. 2020 г.

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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!!) ]
прочитайте больше ..

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CPhill 14 окт. 2019 г.

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'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" прочитайте больше ..

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CPhill 21 окт. 2018 г.

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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....!!! прочитайте больше ..

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CPhill 13 мар. 2018 г.

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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 мар. 2018 г.

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I'm Out Until This Afternoon...You Kids Have Fun !!!

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CPhill 13 февр. 2018 г.

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HAPPY BIRTHDAY, HECTICTAR !!!!

You are a real asset to the forum.....!!!!!

Here's your cake.....enjoy.....!!!!
прочитайте больше ..

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CPhill 14 окт. 2017 г.

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Going To The Picnic

This one isn't too difficult....give it a try.....!!!!!

прочитайте больше ..

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CPhill 25 сент. 2017 г.

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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 прочитайте больше ..

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CPhill 23 сент. 2017 г.

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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 прочитайте больше ..

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CPhill 20 сент. 2017 г.

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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??? прочитайте больше ..

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CPhill 19 мар. 2017 г.

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Happy Pi Day..(3.14)....Everybody.....!!!!!

прочитайте больше ..

hectictar ●●●●●●

CPhill 14 мар. 2017 г.

+129899

The diameter is 2

To see this, note that the circumference is given by C = 2πR, or C=2Rπ. But 2R = D, where D is the diameter.

Therefore, C=Dπ

So 6.28 = Dπ

And if we let π = 3.14

Then, 6.28 = D * 3.14

So D = 6.28 / 3.14 = 2

CPhill26 мар. 2014 г.

+129899

Example:

Twice a number = 2n

So, 5 less than twice would be

2n - 5

CPhill25 мар. 2014 г.

+129899

Take a look at this website, Stu. I've found that his explanations are pretty good. He also gives some examples.

http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx

If you have any specific questions, post them to the forum.

Hope this helps.

CPhill25 мар. 2014 г.

+129899

Although I'm not the most "technical" user of the calculator, it will add and subtract fractions (and reduce them to simplest form). It will also convert an "unknown" decimal to a fraction.

Examples

108/357

9/93+8/102

0.1752055660974067

I'm not sure what you might be after, but if you have more specific questions, post them to the forum and we'll see if we can answer them.

CPhill25 мар. 2014 г.

+129899

First, let sin(x) = x

So

3SIN(x)^2-7SIN(x)-3=0

= 3x^2 - 7x - 3 = 0

Solving, we have.....

3x^2 - 7x - 3 = 0

This implies that sin(x) = (7 - √85) / 6 and sin(x) = (7 + √85) / 6

The second answer is impossible, since it implies that sin(x) > 1

Solving for the first, we have

asin(((7-sqrt(85))/6))

Hope this helps

CPhill25 мар. 2014 г.

+129899

Bertie

Your method of substituting cos(pi/3) for 1/2 and sin (pi/3) for (√3)/2 is quite ingenious!!!!

I like that MUCH BETTER !!!

CPhill25 мар. 2014 г.

+129899

sqrt(3)cos(x) - sin(x)= 1

Let's see if this works

(add sin(x) to both sides and square both sides)

3(cos(x)^2 = 1 + 2sin(x) +sin(x)^2

(using cos(x)^2 = 1 - sin(x)^2, we have)

3*(1 - sin(x)^2) = 1 + 2sin(x) +sin(x)^2

(Simplify)

4*sin(x)^2 + 2*sin(x) - 2 = 0

(Divide through by 2)

2*sin(x)^2 = sin(x) -1 = 0

(Factor)

(2sin(x) -1) (sin(x) +1 ) = 0

Solving, we get x = (pi/6 + 2*pi*n , 3*pi/2 +2*pi*n)

Another 2nd quad angle, 5*pi/6, solves (2sin(x) -1 = 0), but it doesn't "work in the original problem) (due to "squaring," we get an extraneous solution??)

I think that may be it.....but check my work and method!!

CPhill25 мар. 2014 г.

+129899

This is an interesting one........

For all real numbers x,

√(x)^2 = IxI = x if {x ≥ 0 and -x if x < 0}

So, √(-1)^2 = I-1I = -(-1) = 1

Now, many people assume that the result could be "-1" - due to following faulty "proof."

Step 1 √(-1)^2 =

Step 2 √(-1) * √(-1) =

Step 3 ( i * i ) =

Step 4 ( -1 )

The problem comes in the transition from Step 1 to Step 2.

In general, √(a * a) = √(a) * √(a) only if a ≥ 0. (Thus, we can't "split" the radicals in this manner.)

I think that's it !!!

CPhill22 мар. 2014 г.

+129899

How do you Solve this?
2(x+3) - 2x^2(x+3) = 0
In my answer sheet they rewrite it to (x+3) * (2-2x^2) = 0
Im unsure what happens to the other (x+3)

They (almost) took the greatest common factor, (x + 3), out of both terms......... !!!!

We can also take a "2" out !!!

Doing this, we get

2 * (x + 3) * (1 - x^2) = 0

We can get rid of the 2 by dividing through, so this becomes

(x + 3) * (1 - x^2) = 0

To solve, set each factor = 0

The first one is easy

x + 3 = 0

Subtracting 3 from both sides, we get

x = -3 and that's one solution

Now, setting the other factor to 0 we get

1 - x^2 = 0

This can be factored as

(1 - x) * (1 + x) = 0

And setting these two things to 0 (as before), we get the other two solutions of 1 and -1

Hope this helps

CPhill22 мар. 2014 г.

+129899

how do i do a problem like this 1/2(w-4)??

I assume you just want to "distribute" the 1/2 over the terms in the parentheses.

This would just give us

1/2w - 2

CPhill21 мар. 2014 г.

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