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https://web2.0calc.com/questions/probability_27845

I don't see a figure posted, but this problem probably involves the pythagorean theorem

Thank! :)

We can use casework to solve this problem. 

If 3 boxes are empty, then there are two possibilites. 

First, we could have 2 boxes with balls in them. 

Case 1 

Let's say one box and has 7 and the other has 1. 

There are 2 ways we can do this. 

Case 2

Now, let's say there is one boy with 6 and one box with 2, 

There are also 2 ways we can do this. 

Case 3

One box has 3 and the other has 5. 

Another 2 ways we can do this. 

Case 4

One box has 4 and the other has 4. 

Only one way to do this. 

\(2 * 2 * 2 * 1 = 8 \) ways we can do this in total

We choose 2 boxes out of the 5 we could have chosen from, which is just \(5 \choose 2\) = 10. 

\(10 * 8 = 80\) ways to do 2 boxes. 

Now, let's say only 1 box has all the balls. There are 5 ways to choose a box with all 8 balls in it. 

\(80+5=85\). There are 85 ways we could do this!

Thanks! :)

We can use the information we already know to solve this problem!

First off, let's note that \((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\), which we can simplify to \(8=5+6xy\). Solving for xy, we get \(xy=1/2\). 

This may seem useless, but it will come into play later. 

Now, we know that \((x+y)^2 = x^2+2xy+y^2\), which contains x^2+y^2. Isolating this, we get \(x^2+y^2 = (x+y)^2 - 2xy\). 

We already know all the terms! We can plug in 1/2 and 2 which we know from above! We get \(x^2 + y^2 = 4 - 1 = 3\). 

So 3 is our answer!

Thanks! :)

First off, note that \((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\). 

We already know most of the terms, so we have \(8 = 5+3xy(x+y)\), which is the same as \(8=5+6xy\). Now, we have \(xy = 1/2\). 

You may be wondering why in the world would we want xy, but it will make sense later. 

Now, when we see x^2+y^2, (x+y)^2 instantly comes to mind. 

We know that \((x+y)^2 = x^2+2xy+y^2\). This means that \(x^2+y^2 = (x+y)^2 - 2xy\)! We already know all these terms! We can easily plug in 2 and 1/2 to solve our problem!

\(x^2 + y^2 = 4 - 1 = 3\). 

Our final answer is just 3. 

Thanks! :)

        1                               1                           1

_____________  +  _____________ + _______________

sqrt 36 +sqrt 39       sqrt 39 + sqrt 40      sqrt 40 + sqrt 45

Multiplying numerartors/denominators by the conjugate of the  denominators, we can write

 sqrt 36 - sqrt 39         sqrt 39  -sqrt 40       sqrt 40 -sqrt 45

_______________  +  ______________ +  ______________

     36 - 39                        39 - 40                     40 - 45

sqrt 39 - sqrt 36           sqrt 40 - sqrt 39           sqrt 45 - sqrt 40

_____________  +   _______________ +   ______________

          3                              1                                   5

(-2/3)sqrt 39 - 2  + (4/5)sqrt 40  + (1/5)sqrt 45

(-2/3)sqrt 39 -  2 + (4/5)* 2 sqrt 10  + (1/5) 3sqrt 5

(-2/3)sqrt 39  - 2  + (8/5)sqrt 10 + (3/5)sqrt 5

3(2c + 1) = 28*(3c) - 9 + 45(3c) - 18

6c + 3 =  84c + 135c - 27

30  =  213c

c =  30 / 213 =  10  / 71

https://web2.0calc.com/questions/frustum_2

https://web2.0calc.com/questions/plane-and-sphere#r1

sin theta =  [sqrt (4^2 -1^2) ] /  4  =  sqrt (15) / 4

tanx + sec x =  0

sin x / cos x  +  1 / cos x =   0

sin x / cos x =   -1  / cos x           square both sides

sin^2 x / cos^2x  = 1 / cos^2 x

[ 1 - cos ^2 x ]  / cos ^2 x  = 1/ cos^2 x

1/ cos ^2 x  -  1  = 1/ cos^2 x

-1  =  0         →  no solution

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

(3 + 2 + 2 ) !  = 7 !  = 5040 ways

https://web2.0calc.com/questions/triangles_144

First, we can graph this problem. 

We can use the Pythagorean Theorem to complete this problem. 

First, we need similar triangles. We can extend AB towards the direction of A and meet it up with the line connecting the radiuses. We can call this point O. 

Let the center of the unit circle be N and the center of the larger circle M. Let's let x be the distance from the left edge of the smaller circle to O. 

Notice that triangles BMO and ANO are similar triangles. This means we can write

\(BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3 \\ \\ OA = \sqrt {NO^2 - NA^2 } = \sqrt{(1 + 2/3)^2 - 1^2} = \sqrt{25/9 -1 } = \sqrt{ 16/9} = 4/3 \\ OB = \sqrt {MO^2 - MB^2 }= \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt {400/9 -16} = \sqrt{ 256/9} = 16/3 \\ AB = OB - OA = 16/3 - 4/3 = 12/3 = 4 \)

You can also use translation for this problem, but my small brain can't figure it out!

Thanks! :)

I'm not exactly sure what the problem is asking, but I'll but the area in terms of the things we have. 

The volume of the pyramid is 1/3 of the base area mutliplied by the height. 

So we have the area of the pyramid as \(\frac{1}{3}[ABCD]h\) where h is the height. We could probably figure out the height by using the paythagorean theorem through the slant height and the diagonal of ABCD!

I hope I answered your question!

Thanks! :)

https://web2.0calc.com/questions/semicircle_5

Let P and Q  lie on a unit circle

The equation of the line through OP  is   y= 4x

The equation of the line through OQ is y = 5x

To find the x coordinate of P

x^2 + (4x)^2   =1

17x^2 =1

x= 1/sqrt 17      y = 4/sqrt17

To find the x coordinate of Q

x^2 + (5x)^2 =1

26x^2 =1

x = 1/sqrt 26     y =5/sqrt 26

To  find  the slope of the line through  PQ

[ 5/sqrt 26 - 4/sqrt 17 ] / [ 1/sqrt 26 - 1/sqrt 17 ]  =

[ 5 sqrt 17 - 4sqrt26 ] / [ sqrt 17 - sqrt 26 ]  =

[4sqrt 26 - 5sqrt 17 ] / [ sqrt 26 -sqrt 17]

[ sqrt 26 + sqrt 17 ] [ 4sqrt 26 - 5sqrt 17] / 9  =

[ 4 *26 + 4sqrt442 -5sqrt 442 - 85] / 9  =

[19 -  sqrt 442 ] / 9

The slope of the line we want  has the negative reciprocal

9 / [ sqrt 442 - 19 ]  =

9 [ sqrt 442 + 19 ] / [ 442 - 361]

 [sqrt 442 + 19]  /  9  ≈ 4.448

Sum of all edges  =  4[ l + w + h ] = 17 →  l + w + h = 17/4     square both sides

l^2 + w^2 + h^2  + 2 ( lw + lh + wh) = 289/16       (1) 

Surface area  =   2 [ lw + lh + wh ] =  15     (2)

Sub (2)  into (1)

l^2 + w^2 +  h^2  +   15 =  289/16

l^2 + w^2 + h^2  =  49/16

The length of the diagonal  = 

sqrt [ l^2 + w^2 + h^2 ]  =  sqrt [ 49/16 ]   = 7 / 4  =   1.75

Let P = (x,y)

PA^2 + PB^2 + PC^2   =

(x-7)^2 + (y + 11)^2  + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2    .....simplifying, we have

3 x^2 - 70 x + 3 y^2 + 40 y + 1247

3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247       complete the square on  x,y

3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12

3 [ (x - 70/6]^2 + (y + 40/6)^2 ] +  2116 / 3

Q = (70/6, -40/6 ) =  (35/3 , - 20/3)

k = 2116 / 3

First of all, let's move almost all the variables to one side of the equation and combine all the like terms. 

We get \(x^2 - 21x + y^2 + 13y = 25 \). 

Now, we complete the square for both x and y on the right side of the equation. 

\(x^2 - 21x + 441/4 + y^2 + 13y + 169/4 = 25 + 441/4 + 169/4 \)

\((x - 21/2)^2 + (y + 13/2)^2 = 355 / 2 \)

Wait! This looks like the equation for a circle! 

Indeed, we have a circle at center \((21/2 , -13/2) \) with radius \(\sqrt{355 / 2}\). 

The value of the largest possible x is just the radius added onto the x value of the center. 

\( x = (21/2) + \sqrt{355 / 2}\)

Thanks! :)

1. The answer is 24 degrees.

2. The number of possible locations is 4.

3. x = 56 degrees.

The value of x is 3*sqrt(2).

The rope is 24 m long.