All Answers 356344 Answers

Guest's answers are not correct.

I will start with an equilateral triangle inscribed in a unit circle (radius=1)

Using cosine rule it can be determined that the side length of the triangle is sqrt3

Choose any first point and call it A

The second point can be any point up to sqrt3 units left or right of A so that is a total of 2sqrt3 units

the circumference is 2pi units

So the prob that the 2 points are  closer together  than the sides of the triangle is     

\(\frac{2\sqrt3}{2\pi}=\frac{\sqrt3}{\pi}\)

why is is this nig-ger sh-it on here

In order to find the range, 

Let \(y=r(x)\)

      \(y={x^2 \over 1-x^2}\)

⇒  \({1\over y}={1\over x^2}-1\)

⇒  \({1\over x^2}={1+y\over y}\)

⇒  \(x^2={y\over 1+y}\)                          ...(1)

Eq (1) has solutions if and only if, 

\({y\over 1+y} ≥0\)

i.e. if either \(y≥0\)  and  \(y+1>0 \)

or  \(y≤0\)  and  \(y+1<0\)

Thus range of r(x) is \( ( − ∞ , − 1 ) ∪ [ 0 , ∞ )\).

~Thank You!

Thanks a lot Phill! 

Say that the first one does happen.  What are the probabilities that the others 2 won't happen?  (Do them seperately)

Multiply the three probs together.

Then do the same for the second one.  etc

when you have done all three, add your answers together.

This is a contour probability map.

The shaded region is the desired region.  

The prob is the area of it over the area of the entire region.

A+B+C can only equal  6 or 12 base 10    ( because 3 whole numbers less than 6 can add at the most to 15)

If A+B+C=10 base 6,  then A=1    The sum of B and C is easy to see.

If A+B+C=20 base 6,  then A=2   The sum of B and C  is easy to see.

One of these sums can be discounted immediately. 

Bingo you have your answer.

Good to be back! Albeit only on rare occasion, since I'm not in school anymore (had to drop out due to finances). And I haven't taken a maths course (to completion) since... gosh, 2016? But I'll still be here to keep my mind sharp, and to stay learned!

Since 60 degrees is 1/6 of the circle = 360 degrees, the probability is 1/6.

Let's write each integer from 0 to 599 with 0-padding to 3 digits.  So they are 000, 001, 002, ..., 599.  Let $S$ be the set of integers from 0 to 599, inclusive, that contain neither the digit 5 nor 6.   Then the answer we are seeking is $601-|S|$.  (There are 601 numbers from 0 to 600 inclusive; the number 000 is in $S$; and the number 600 contains 6 at least once.)

Looking at it this way, numbers in $S$ can be constructed like this:  Firstly, choose the leftmost digit from 0, 1, 2, 3, or 4.  There are 5 choices.  Secondly, choose the middle digit from 0, 1, 2, 3, 4, 7, 8, 9.   There are 8 choices.  Lastly, choose the rightmost digit from 0, 1, 2, 3, 4, 7, 8, 9.   Again there are 8 choices.   This gives $5\cdot 8\cdot 8= 320$ numbers in S.

So the answer to our question is $601-320=281$.

100a + 10b + c

a, b, c is a decreasing arithmetic sequence. 

a, b, c + 2 is a geometric sequence.

Note that the c can't carry over since that only happens when c is 9 (odd number) or 8, but 8 + 2 = 0 which can't be part of a geometric sequence. 

c = c

b = c + x

a = c + 2x

b^2 = (c + 2)(a)

(c + x)^2 = (c + 2)(c + 2x)

c^2 + 2cx + x^2 = c^2 + 2cx + 2c + 4x

x^2 = 2c + 4x

Let's try when c = 2. 

x^2 = 4 + 4x, x is a weird number. 

Let's try when c = 4. 

x^2 = 8 + 4x, x is not a nice number. 

Let's try when c = 6.

x^2 = 12 + 4x, x is a nice number. 

(x-2)(x-6) = 0

x = 6, this is not possible. 

Let's try when c = 0. 

x^2 = 4x

x = 4. 

c = 0

b = 4

a = 8

Our number is 840. :))

=^._.^=

PLS GIVE ME MANY HEARTS THIS WAS VERY HARD TO TYPE :c

+119736 CPhill moderator

+113548 Melody moderator

+32541 ElectricPavlov

+32236 Alan moderator

+25911 heureka

+22081 geno3141

+12511 Omi67 moderator

+11607 asinus moderator

+9292 hectictar

+8460 MaxWong

+5261 rarinstraw1195

+4164 zegroes

+3692 BrittanyJ

+3111 TitaniumRome

+3104 adminadministrator

+2555 Solveit

+2181 CalTheGreat

+2169 GingerAle

+2167 Probolobo

+1746 catmg

+1491 HighSchoolCalculus

+1467 Dragan

+1436 SmartMathMan

+1301 civonamzuk

+1291 jugoslav

+1250 CoolStuffYT

+1195 GAMEMASTERX40

+1164 jimkey17

+1004 GoldenLeaf

+956 CubeyThePenguin

+935 dgfgrafgdfge111

+890 Bertie

+870 EinsteinJr

+862 AnimalMaster

+846 juriemagic

+762 whymenotsmart

+760 CentsLord

+734 lokiisnotdead

+656 amazingxin777

+564 chilledz3non

+557 iamhappy

+555 thedudemanguyperson

+554 xxJenny1213xx

+544 Logarhythm

+536 MooMooooMooM

+528 Badinage

+514 mathismyhobby

+506 SparklingWater2

+500 ellapow

+484 DarkCalculis

+484 RiemannIntegralzzz

+482 MathProblemSolver101

+470 Neptune

+468 Mr.Owl

+421 APatel

+421 Pangolin14

+408 liz54321

+402 ss333

+394 MobiusLoops

+365 kes1968

+362 beababy

+362 imdumb

+354 textot

+353 Mathefreaker2021

+337 mworkhard222

+336 wolfiechan

+332 Dennis070sinneD

+326 TheMobMaster2006

+326 KennedyPape

+325 Varxaax

+321 itachiuchihakujo

+311 thelizzybeth

+311 WillBillDillPickle

+308 weredumbmaster22

+307 amygdaleon305

+295 IsabelleJ761

+294 calvinbun

+276 Mathisgood

+270 Tgfromthe4

+268 IhaveOBJD

+267 KitSoundwave

+266 DewdropDancer

+258 penquino21

+255 wiseowl

+249 JonathanB

+246 power27

+237 BWStar

+236 Choff

+236 Charger421

+232 ramenmaster28

+224 TaliaArticula

+222 CPhilFanboy

+221 username

+219 EmeraldWonder

+218 AvenJohn

+217 Oofrence

+212 OldTimer

+211 Noori

+207 lhyla02

+205 TacoBell

+203 TheEarlyMathster

To look at this problem more clearly, let's write it in latex... $1-1x+2x^2-2-3x+3x^2-4+4x-5x^2-6x+7-8x^2$

We can easily combine like terms to get $(2+3-5-8)x^2 + (-1-3+4-6)x + (1-2-4+7) = -8x^2 -6x + 2$.

Can you finish from here? 

Since $P(x)$ is cubic and $x^2+x+1$ is quadratic, the quotient must be linear.   Let the quotient be $ax+b$ for some constants $a$ and $b$.
Then
$$P(x) = (x^2+x+1)(ax+b) + 2x+3.$$
We are given that $P(0)= -3$, so
$$(0^2+0+1)(a(0)+b) + 2(0)+3 = -3$$
that is,
$$b+3=-3,$$
giving $b=-6$.
We are also given that $P(1) = 4$, so
$$(1^2+1+1)(a(1)-6) + 2(1)+3 = 4$$
that is,
$$3(a-6)+5=4$$
giving $a=\frac{17}{3}$.
Therefore, the quotient is $\frac{17}{3}x-6$.

thank you daddy cphill!

0.05 * 4 = 0.20

1.00 - 0.20 = 0.80

80 members in the club

x/3 = x/4 + 18

-x/4     -x/4

(x/3 - x/4) = 18

*multiply to change the denominators*

(4x/6 - 3x/6) = 18

x/6 = 18

6(x/6) = 6(18)

x = 108

Now we just halve that to get the number

108 ÷ 1/2 = 54

54 is the answer

4 boys   must represent  5%  of  the club

So....the total members  must  be  4 / 5%    4/.05  =    =   80

The slope of two points is $\frac{y_1-y_2}{x_1-x_2}$. In order for the slope to be $\frac{1}{7}$, this equation has to equal $\frac{1}{7}$. 

Plugging in values gives $\frac{4-y}{-3-4} = \frac{1}{7}$. Solving this equation gives $y = \boxed{5}$.