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6  9  12  16    looks like    

arithmetic sequence   d = 3

geometricsequence   r = 1 1/3

x=9 y=12

6x + 7  =  11x  -  13      add 13 to both sides, subtract 6x from both sides

20  =  5x           divide both sides by 5

4  =  x

If sin x + cos x = -1/5, and 3/4*pi <= x <= pi, then find cos 2x.

sin x  +  cos x  =  -1/5        square both sides

sin^2 x + 2sinxcosx  + cos^2x  = 1/25

1 + 2sin xcosx =  1/25

2sinxcosx =  -24/25

sin (2x)  =  -24/25

cos (2x)  = ±√ [ 1 - sin^2(2x)  ]

cos (2x)  = ±√[ 1 - (-24/25)^2 ] 

cos (2x)  = ±√[ 625 - 576]/ 25

cos (2x)  =  ± (7/25)

Since  x =    [ (3/4)pi , pi ]   then

arccos [ 7/25]  = x  ≈  1.287 rads

Then dividing by 2, x  ≈  .6435 rads......reject

arccos [ -(7/25) ]   = x  ≈  1.85 rads

Then dividing by 2, x    ≈  .927 rads   .....reject

And

arccos  [ 7/25]  =  2pi - 1.287  =  4.996 rads

Then dividing by 2, x ≈  2.498 rads.....accept

So cos 2x   =  7/25

Also can you explain how you got the probability of P(n)?

Is n the number of balls you make?

Angle HCA = 44 degrees.

The altitude and the median from vertex A of triangle ABC are 4 and 5 units long, respectively.  The anlge determined by AB and the median is bisected by the altitude.  Find AC.

Did you know that if you uploaded a picture or diagram of you problem, that you will get a much quicker response to your question than without it ??

oops ignore the abouve one, 

x+8-4(x-1)=2x+3

x+8-4x+4=2x+3 

add 4x 

x+8+4=6x+3

x+12=6x+3

subtract x

5x+3=12

subtract 3

5x=9 

x=9/5=1.8

x+8-4x+4=2x+3 

Combine like terms 

-3x+12=2x+3

-x=-9 

x=9 

x+y = a (1)

\(x^3+y^3=b\) (2)

\(x^3+y^3=(x+y)(x^2-xy+y^2)\) (4)

\((x+y)(x^2-xy+y^2)=b\) (3)

\(\frac{b}{(x+y)}=(x^2-xy+y^2)\)

From (1) we know that x+y=a

so

\(\frac{b}{a}=x^2-xy+y^2\) Expressed in terms of a and b as the question asks. and a can't be 0 as given. 

\(x^2+y^2=\frac{b}{a}+xy\) Expressed in terms of \(x^2+y^2\)

Notice that 

\((x+y)(x^2-xy+y^2)\)

we know \(x^2+y^2=\frac{b}{a}+xy\)

we can subsituite 

we also know (x+y)=a 

so

\(a*(\frac{b}{a}+xy-xy)\)

Simplify the brackets

\(a*(\frac{b}{a})\)

which is equal to b ,=x^3+y^3 as given.

So this identity is correct that x^2+y^2=b/a+xy

Here is a graph that might clue ya...

https://www.desmos.com/calculator/y0gwcmkik7

The oint C divides the line in to two segments   one is 3/7   the other is 4/7

  we want to find the point that is 3/7 from point A

The distance between the x coordinates   from -1 to 2 is 3 units     3 units x 3/7 = 9/7   add this to -1

-1 + 9/7 = 2/7   x coordinate

The distance between y coordinates    from 4 to -3   is -7 units     -7 x 3/7 = -3  add this to 4

4 - 3 =  1   y coordinate

Here is a graph:

https://www.desmos.com/calculator/zqpuoy5cik

W = F x d     Work = Force x distance

F = ma    so work becomes    W = ma x d

W = 500 kg (9.81 m/s²⁾  x 2 m

W = 500(9.81)(2) joules

PolynomialGCD[12 x, 3 x^2 + 6 x] = 3x

A) The number of ways is 6! = 720.

B) The number of ways is 5! = 120.

C) The number of ways is 6*4! = 144.

1. We can count that there are 3*C(7,5) = 63 quadrilaterals that works, so the probability is 63/(4*C(11,4)) = 21/440.

2. The cases are where we draw 4 pennies and 0 nickels, and 1 nickel and 3 pennies.  This leads to a probability of (6*5 + 18*10)/C(11,4) = 7/11.

3. We must consider strings like BBB and BBTB.  Going through the cases, we get a probability of (2/3)^3 + 3*(2/3)^3*(1/3)^2 + 6*(2/3)^3*(1/3)^4 = 304/729.

We can solve each inequality, which gives us the interval (-7/3, 18].

Hows school?

Hello :)

\(\text{Let }x = \sqrt{10 +\sqrt{10 +\sqrt{10 + \cdots}}}\\ x = \sqrt{10 + x}, x \geq 0\\ x^2 = 10 + x\\ x^2 - x - 10 = 0\\ x = \dfrac{1\pm\sqrt{1^2 - 4(1)(-10)}}{2}\\ x = \dfrac{1\pm\sqrt{41}}{2}\\ \text{Rejecting the negative root, we have }\boxed{x = \dfrac{1+\sqrt{41}}{2}} \)

Observing the recurrence relation, \(\{b_n\} \) is a quadratic sequence.

Let \(b_n = pn^2 + qn + r\qquad p,q,r\in \mathbb R\)

Obviously, \(a_n = 4 + 2n\).

\(b_1 = 3\\ p(1)^2 + q(1) + r = 3\\ p + q +r = 3 \phantom{-----}(1)\)

\(b_n = b_{n - 1} + a_{n - 1}\\ pn^2 + qn + r = p(n - 1)^2 + q(n - 1) + r +(4+2(n-1))\\ pn^2 +qn + r = pn^2 - 2pn + p + qn - q + r + 4 + 2n - 2\\ 0=-2pn+p-q+2+2n\\ (2 - 2p)n+(p-q+2)=0n+0\\ \text{Comparing both sides,}\\ 2-2p = 0\\ 2p = 2\\ \boxed{p = 1}\\ p -q+2=0\\ 1-q+2=0\\ \boxed{q=3}\\ \text{Substitute }p=1\text{ and }q = 3\text{ into equation }(1),\\ 1+3+r=3\\ \boxed{r = -1}\\ \therefore b_n = n^2 + 3n - 1\)