All Answers 358516 Answers

Hi Julio

It is 12:05am

I suppose that is morning. :)

Anonymous4338 almost got it right....except the slope is MINUS 1/3 .

Slope = rise/run  (when looking at your points from L to the Right one)

From the two points    rise = 7-10 = -3     run = 6- (-3) = 9     Rise/run = -3/9 = -1/3

Now y=mx+b   Selecet one of your set of points, say  6,7

7=(-1/3)6 +b

b= 9

SO the eqution for your line is    y= (-1/3)x + 9

Test this by substituting in your OTHER point

10 = (-1/3)(-3) + 9

10=1+9 = 10             It works!

~jc

Thanks Heureka,

That excursion was really convoluted.

Intriguing - but convoluted.    

If that is 26 inch diameter then

circumference = pi * diameter

~3.14*26

~81.64 inches

Thurs 10/12/15

If lnx=1/x, then what is ln1/x=?    Thanks Heureka

http://web2.0calc.com/questions/lnx-1-x--gt-ln1-x

Tips for solving trig equations    Thanks Melody 

http://web2.0calc.com/questions/any-tips-of-solving-simplify-verify-and-prove#r1

What is a geometric mean   Thanks Heureka

https://web2.0calc.com/questions/help-with-geometric-means

Finding the nth term   Thansk CPhill, Heureka and guest.   :)

http://web2.0calc.com/questions/find-the-nth-term-of-this-sequence

Given sine find cos        Thanks Melody and Will85237

https://web2.0calc.com/questions/i-need-a-little-help

What kind of regular polygon is it?  Thanks Melody and Hayley.

https://web2.0calc.com/questions/geometry_79609

Using a Venn diagram to help with a probability question  Thanks CPhill

https://web2.0calc.com/questions/i-really-need-hep-for-that-please

Burning the candle at both ends.   Thanks Heureka

https://web2.0calc.com/questions/burning-the-candle-at-both-ends

Rationalizing the denominator when it is a cube root.   Melody

https://web2.0calc.com/questions/y-7x-4-sqrt3-2-7-x-4-sqrt3-2

@@ What is Happening?  [Wrap4]   Wed 9/12/15   Sydney, Australia Time 11:26 pm   ♪ ♫

Hi all,

Answer credits go to CPhill, Alan, Heureka, EinsteinJr, Riddle, DragonSlayer554, James10898, Omi67, Nauseated, GoldenLeaf, Anonymous4338, TijustaGod, and SpawnofAngel.  Thank you.  

There may well be others as well, I am sorry if you got omitted.  The forum is too big and I have been too busy to really know what is going on    If you answer a question and you believe your answer is correct and reasonably well explained then I strongly advise you to give yourself 5 stars.  You can no longer rely on me to give the points to you.  I do not see all posts anymore  

Technical Issues

IDK, is the picture upload working - it wasn't the last time i tried it. ://

How can you post a picture when the picture upload on the forum is not working properly ?

Thanks very much Alan for these instructions.

http://web2.0calc.com/questions/ven-diagram-need-help-for-questions#r3

Interest Posts 

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

These interest posts are not a good cross section.  Sorry.

1)     Riddle      Thanks DragonSlayer554

if lnx=1/x, then what is ln1/x=?

\(\begin{array}{rcll} \ln{(x)} =\frac{1}{x} \qquad \ln{( \frac{1}{x} )} = \ ? \end{array}\)

\( \begin{array}{rcll} \ln{( \frac{1}{x} )} &=& \ln{( 1 )} -\ln{( x )} \qquad & | \qquad \ln{( 1 )} = 0\\ &=& 0 -\ln{( x )} \\ &=& -\ln{( x )} \qquad & | \qquad \ln{(x)} =\frac{1}{x}\\ &=& -\frac{1}{x} \\ \mathbf{ \ln{( \frac{1}{x} )} } & \mathbf{=} & \mathbf{ -\frac{1}{x} } \end{array}\)

excursus

\(\begin{array}{rcll} \ln{(x)} = \frac{1}{x} \qquad x = \ ? \end{array}\)

\(\begin{array}{rcll} \ln{(x)} &=& \frac{1}{x} \\ \ln{(x^x)} &=& 1 \qquad & | \qquad e^{()} \\ \mathbf{x^x} &\mathbf{=}& \mathbf{e} \\\\ e^{\ln{(x^x)}}&=& e \qquad \ln{(x^x)} = 1\\ e^{ x\cdot \ln{(x)} } &=& e\\ x\cdot \ln{(x)} &=& 1 \qquad z = \ln{(x)} \\ x\cdot z &=& 1 \qquad e^z = x \\ e^z\cdot z &=& 1 \\\\ z &=& W(1) \\ e^{\ln{(x)}} &=& e^{W(1)}\\ x &=& e^{W(1)} \qquad \text{or}\\\\ x\cdot z &=& 1 \\ x &=& \frac{1}{z} \qquad z = W(1) \\ x &=& \frac{1}{W(1)}\\\\ \end{array}\\ \small{ W(1) = 0.5671432904097838729999686622103555497538157871865125081351310792230457930866\dots\\ x = \frac{1}{W(1)} = 1.7632228343518967102252017769517070804360179866674736\dots\\\\ W(1) = 0.5671432904\dots \text{ is called the omega constant }\\ \text{where } W(x) \text{ is the Lambert W-function } }\)

g(x)=x when 0<x<1

This is an interval but it is not a question.  ://

If it were graphed then

It is just an interval from (0,0) to (1,1) where the ends are not included.

how to write 3/5 to 6 fraction in simplest form?'

\(\frac{3}{5}:6\\~\\ 5*\frac{3}{5}:5*6\\~\\ 3:30\\~\\ 1:10 \)

You don't say what you are trying to calculate, so I've assumed you want to find the approaching velocities.

Also you can't have a mass in Newtons, so I've converted your 6500 pounds (mass) to kg.  (there is a difference between lbf (force) and lb (mass)).

On second thoughts I should probably have assumed kinetic friction.  However, static friction is usually higher than kinetic friction, so perhaps I've used the right numerical value after all!

Because it only has one factor, to be prime the number must have two factors.

so that means that the first 4 are right handed so...
0.86 * 0.86 * 0.86 * 0.86 * 0.14 = 0.077
being that the probability of a right handed person being selected is 0.87 and left handed is 0.14

you need to purchase at least 12 plants to landscape your front yard. Trees cost $30 and bushes cost $8. Let "x" represent the number of trees, and let "y" represent the number of bushes. You can't spend more than $200. If you purchase 2 trees, how many bushes could you buy?

The two trees will cost 2*30 = $60

So you have 200-60 = $140 left for bushes

140 divided by 8 = 17.5

you cannot buy half a bush so you can buy 17 bushes.

(-2) times (-2)=2Times 2=4

17.5trees

Given a triangle with a=11, b=14, and \(\mathbf{\alpha = 15^{\circ}}\), what is (are) the possible length(s) of c?

\(\small{ \begin{array}{rcll} a^2 &=& b^2+c^2-2bc\cdot \cos{(\alpha)} \\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \\\\ \boxed{~ \begin{array}{rcll} Ac^2+Bc+C = 0 \\ c = {-B \pm \sqrt{B^2-4AC} \over 2A} \end{array} ~}\\\\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \qquad A=1 \qquad B = -2b\cdot \cos{(\alpha)} \qquad C = b^2-a^2 \\ c &=& {2b\cdot \cos{(\alpha)} \pm \sqrt{(2b\cdot \cos{(\alpha)})^2-4(b^2-a^2)} \over 2} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ [ b\cdot \cos{(\alpha)} ]^2-(b^2-a^2)} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ \cos^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ 1-\sin^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ a^2 - b^2\cdot \sin^2{(\alpha)} } \quad a=11 \quad b=14 \quad \alpha = 15^{\circ}\\ c &=& 14\cdot \cos{(15^{\circ})} \pm \sqrt{ 11^2 - 14^2\cdot \sin^2{(15^{\circ})} } \\ c &=& 14\cdot 0.96592582629 \pm \sqrt{ 107.870489571 } \\ c &=& 13.5229615680 \pm 10.3860719028\\\\ c_1 &=& 13.5229615680 + 10.3860719028\\ c_1 &=& 23.9090334709\\\\ c_2 &=& 13.5229615680 - 10.3860719028\\ c_2 &=& 3.13688966521 \end{array} }\)

check:

\(\begin{array}{rcll} c_1\cdot c_2 &=& (b-a)(b-a+2a)\\ c_1\cdot c_2 &=& (b-a)(b+a)\\ c_1\cdot c_2 &=& b^2-a^2 \quad a=11 \quad b=14 \\ c_1\cdot c_2 &=& 14^2-11^2 \\ c_1\cdot c_2 &=& 75 \\\\ c_1\cdot c_2 &=& 23.9090334709\cdot 3.13688966521 \\ c_1\cdot c_2 &=& 75\quad \text{ Okay}\\ \end{array}\)

the expression (3/2x+1)(3/2x-1)-(3/2x-1)^2 is equivalent to?

\(\begin{array}{ll} \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)^2 \\\\ = \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)\left( \frac32x-1 \right) \qquad & | \qquad \text{factorise } \left( \frac32x-1 \right) \\\\ = \left( \frac32x-1 \right) \left[ \left( \frac32x+1 \right) - \left( \frac32x-1 \right) \right]\\\\ = \left( \frac32x-1 \right) \left( \frac32x+1 - \frac32x+1 \right)\\\\ = \left( \frac32x-1 \right) \left( 2 \right)\\\\ = 2 \left( \frac32x-1 \right) \qquad & | \qquad \text{expand} \\\\ = 2\cdot \frac32x-2\cdot 1 \\\\ = 3x-2 \\\\ \mathbf{ \left( \frac32x+1 \right) \left( \frac32x-1 \right) - \left( \frac32x-1 \right)^2 } \mathbf{ = 3x-2 } \end{array}\)

tan^2(30)= 1/cos^2(30) - 1 explain why. 

\(\begin{array}{rcl} \text{Use the Pythagorean identity } \sin^2{(\theta)} + \cos^2(\theta) =1 \\ \end{array}\)

\(\begin{array}{rcl} \sin^2{(30^{\circ})} + \cos^2(30^{\circ}) &=& 1 \qquad & | \qquad : \cos^2(30^{\circ})\\\\ \frac{\sin^2{(30^{\circ})} } {\cos^2(30^{\circ})} + \frac{ \cos^2(30^{\circ}) } {\cos^2(30^{\circ}) } &=& \frac{1} { \cos^2{(30^{\circ})} } \qquad & \frac{\sin^2{(30^{\circ})} } {\cos^2(30^{\circ})} = \tan^2{(30^{\circ})}\\\\ \tan^2{(30^{\circ})} + 1 &=& \frac{1} { \cos^2{(30^{\circ})} } \qquad & | \qquad -1\\\\ \mathbf{\tan^2{(30^{\circ})} }& \mathbf{=} & \mathbf{ \frac{1} { \cos^2{(30^{\circ})} } -1} \end{array}\)

Good spotting Guest 4     

I really need to wipe the mud off my glasses.    

One ton is equal to 2000 pounds, so let's do 2000*1 1/5.

2000*1 1/5=2400

So, 1 1/5 tons = 2400 lbs

There are literally infinite numbers that add up to make -10, but a list of whole numbers that add up to -10 are...

0 and -10

-1 and -9

-2 and -8

-3 and -7

-4 and -6

-5 and -5

However, there is an infinite number of other possibilities (such as -40 and 30 or -1.2 and -8.8)

Oh, and @Melody there's a general gold standard that is used through the world, and the price of it is tracked.

And since I apparently can't edit my older post, all currency values are in US Dollars (USD). If you wanted to find the value in other currencies, simply convert the final value ($878,144.38 USD) from USD into whatever your preferred currency is.

To solve simple equations like this, just use the site's calculator by clicking on the Home tab the the top of the page.

(-2)/(-2)=1

Any number divided by itself is 1 except for 0, which has no answer.

i cant tell if you are being sarcastic or not, but i shall explain it anyway. By dividing, you are in a sence just making a fraction, the first number on top and the number you are using to divide on the bottom. It looks a bit like this ; -2/-2 (i cant do it here, but one of the -2 is obove the other). now you look for similarities. the first major one is that they are both divisible by two, so you do that to both, top and bottom. The next that to notice is that the question now looks like this; -1/-1. Since both are negative, they cancel the negative and you are left with 1/1 which equals 1.