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Perimeter =10 + b + c

30 =10 + b + c

20 = b + c

c = 20 - b.......................(1)

By Pythagoras's Theorem:

c^2 =10^2 + b^2.........(2)         sub c from (1) above into (2)

(20 - b)^2 = 10^2 + b^2

400 -40b + b^2 =100 + b^2    subtract 100 + b^2 from both sides

300 - 40b =0                               add 40b to both sides

300 =40b                                     divide both sides by 40

b =300 / 40

b=7.5                                         sub this into (1) above

c =20 - 7.5

c = 12.5 - This is accurate, because:

12.5^2 =7.5^2 + 10^2

156.25 =56.25 + 100

N =(L - F) / D +1, where N=Number of terms, L=Last term, F=First term, D=Common difference.

N=(73 -1) / 2 + 1

N =37 terms

Sum = (F + L) / 2 x N

Sum = (1 + 73) / 2 x 37

Sum =1,369

Do exactly the same thing with numbers 1 to 231. If you don't make any mistakes, you should get:

Sum =13,456

C)

Sum = 1/256 x [ 1 - (-4)^10] / [1 - (-4)]

Sum = 1/256 x [1 - 1,048,576] / [1 + 4]

Sum = 1/256 x [-1,048,575] / [5]

Sum = 1/256 x [ - 209,715]

Sum = - 209,715 / 256

THX! 28 was the answer

Hi Melody

I'm worried about your stars and bars method. I don't think that it works for this sort of problem, (without some sort of tweaking).

Take the situations where the balls are split 6 - 0 - 0, or 0 - 6 - 0, or 0 - 0 - 6.

Label the first pair of bars b1 and b2, the second pair b3 and b4, and so on through to the final pair b13 and b14.

The 6 - 0 - 0 possibility is catered for by the pair b13 and b14, and that pairing is unique.

Similarly 0 - 0 - 6 is catered for by b1 and b2, again unique.

What about 0 - 6 - 0 though ? Are there not four possible pairings ? b1 and b13, b1 and b14, b2 and b13, b2 and b14.

It looks like its counting this single possibility four times.

When I looked at this problem I went back to basics.

The six balls can be split in seven different ways,

6 - 0 - 0,

5 - 1 - 0,

4 - 2 - 0,

4 - 1 - 1,

3 - 3 - 0,

3 - 2 - 1,

2 - 2 - 2. 

The first of these can happen in three different ways, the number of boxes into which the 6 can be put. The second can happen in six ways, the 5 can be put into any of the three boxes and then the 1 into either of the remaining two, etc.

The seven frequencies will be 3, 6, 6, 3, 3, 6, 1, totalling 28 in all.

I looked in at that web site mentioned in the earlier post.

In there it gives the formula, (for k indistinguishable balls and n distinct boxes),  (k+n-1)C(n-1).

That produces the result (6+3-1)C(3-1) = 8C2 = 28.

Tiggsy

Arithmetic sequences are written in the form of the following:

 \(a_n=a_1+d(n-1)\).

a_n is the nth term of the sequence

a₁ is the first term of the sequence 

d is the common difference

n is the desired term number

1) We already know the information necessary to write an equation for the nth term of this sequence. 

2) There is a formula that exists for the summation of an arithmetic series or geometric series, but that probably would not help your understanding anyway; I can derive it for you, though. 

We have to know the last term, and we can generate this by using the formula from before: \(a_n=a_1+d(n-1)\). Therefore, n=10, d=5, and a₁=5:

Now, let's attempt to evaluate the sum. Standard notation dictates \(S_n\) for the summation. 

I did not answer every question here because the others can be answered with the knowledge given above, albeit not directly. Try and figure it out yourself. 

Find the sum: 75 + 77 + 79 + ... 229 + 231.

\(\begin{array}{|rcll|} \hline && 75 + 77 + 79 + \cdots 229 + 231 \\ &=& (75+0) + (75+2) + (75+4) + \ldots + (75+154) + (75+156) \\ &=& (75+0) + (75+1\cdot2) + (75+2\cdot2) + \ldots + (75+77\cdot2) + (75+78\cdot2) \\ &=& 75 + 75 \cdot 78 + 2 \cdot (1+2+\ldots + 77+78) \\ &=& 75 \cdot 79 + 2 \cdot \dfrac{(1+78)\cdot78}{2} \\ &=& 75 \cdot 79 + 2 \cdot \dfrac{79 \cdot 78}{2} \\ &=& 75 \cdot 79 + 79 \cdot 78 \\ &=& 79\cdot(75+78) \\ &=& 79\cdot 153 \\ &\mathbf{=}& \mathbf{12087} \\ \hline \end{array}\)

You have a row of 10 doors (maybe in a wing of a hotel). 

You are going to paint the doors red and blue, but you don't want two red doors next to each other,

(having blue doors next to each other is fine).

How many ways can you pain the row of doors?

\(\begin{array}{|r|r|} \hline \text{red doors in the row} & \text{no two red doors next to each other} \\ \hline 0 & 1 \\ 1 & 10 \\ 2 & 36 \\ 3 & 56 \\ 4 & 35 \\ 5 & 6 \\ \hline \text{sum} & 144 \\ \hline \end{array}\)

You can pain the row of doors 144 ways.

3e+40 = 3.0*10^-40

That site looks more friendly than some others that people take us to but i have not worked through it.

I try not to rely on formulas, they can be misleading, especially with counting theory and probability.

I am nost saying categorically that mine is correct,  it could be wrong, BUT I just like to understand why an answer is true (or not true) and not just rely on the 'magic box' that a forrmula is. 

If you would like to discuss this other answer I am happy to listen :)

how am i solving this geometric series wrong?

\(\begin{array}{|rcll|} \hline && (-4)^{10} \\ &=& {\color{red}+}4^{10} \qquad \text{ because the exponent is even!} \\ &=& 1048576 \\ \hline \end{array} \)

Be really great if someone could answer this :D

Let \(\mathcal{R}\) denote the circular region bounded by x^2 + y^2 = 36.
The lines x = 4 and y = 3 partition \(\mathcal{R}\) into four regions \(\mathcal{R}_1, \mathcal{R}_2, \mathcal{R}_3, \text{ and } \mathcal{R}_4\).
Let \([\mathcal{R}_i]\) denote the area of region\( \mathcal{R}_i\).
If \([\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4]\), then compute \([\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]\).

\(\text{Let $ \mathcal{R}_1 = \mathcal{R}_{11} + \mathcal{R}_{12} + \mathcal{R}_{13} + \mathcal{R}_{14} $ } \\ \text{Let $ \mathcal{R}_2 = \mathcal{R}_{21} + \mathcal{R}_{22} $ } \\ \text{Let $(1)\qquad \mathcal{R}_2 = \mathcal{R}_{13} + \mathcal{R}_{14} $ } \\ \text{Let $(2)\qquad \mathcal{R}_3 = \mathcal{R}_{12} + \mathcal{R}_{13} $ } \\ \text{Let $(3)\qquad \mathcal{R}_4 = \mathcal{R}_{21} $ } \\ \\ \text{Let $(4)\qquad \mathcal{R}_{13} = \mathcal{R}_{21} $ } \\ \text{Let $(5)\qquad \mathcal{R}_{1} -\mathcal{R}_{13}-\mathcal{R}_{14} = \mathcal{R}_{12} + \mathcal{R}_{11} $ }\)

\(\begin{array}{|rcll|} \hline [\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4] &=& ([\mathcal{R}_1] - \underbrace{[\mathcal{R}_2])}_{= [\mathcal{R}_{13}] + [\mathcal{R}_{14}]} - (\underbrace{[\mathcal{R}_3]}_{=[\mathcal{R}_{12}] + [\mathcal{R}_{13}] } - \underbrace{[\mathcal{R}_4]}_{=[\mathcal{R}_{21}] } ) \\\\ &=& ( \underbrace{[\mathcal{R}_1] - [\mathcal{R}_{13}] - [\mathcal{R}_{14}]}_{=[\mathcal{R}_{12}] + [\mathcal{R}_{11}] } ) -( [\mathcal{R}_{12}] + \underbrace{[\mathcal{R}_{13}]}_{=[\mathcal{R}_{21}]} - [\mathcal{R}_{21}] ) \\ \\ &=& ( [\mathcal{R}_{12}] + [\mathcal{R}_{11}] ) -( [\mathcal{R}_{12}] + [\mathcal{R}_{21}] - [\mathcal{R}_{21}] ) \\ \\ &=& ( [\mathcal{R}_{12}] + [\mathcal{R}_{11}] ) -[\mathcal{R}_{12}] \\ \\ &=& [\mathcal{R}_{11}] \\ \\ &=& 8 \times 6 \\ \\ &\mathbf{=}& \mathbf{ 48 } \\ \hline \end{array}\)

This last region is simply a rectangle of height 6 and width 8, so its area is 48.

-339.990234375 / -0.5  x 512/512

174,075 / 256

the textbook answer is \({174075 \over 256}\) as the exact value in fraction form though :")

Sum=6 x [1 - 1.5^10] / [1 - 1.5]

Sum=6 x [1 -  57.665] / [-0.5]

Sum=6 x [-56.665] / [-0.5]

Sum =6 x   -56.6650390625 / - 0.5 

Sum =-339.990234375 / -0.5

The numbers with odd divisors under 400 are as follows:
1, 2^2, 2^4, 2^6, 2^8, 3^2, 3^4, 5^2, 6^2, 7^2, 10^2, 11^2, 12^2, 13^2, 14^2, 15^2, 17^2, 18^2, 19^2=19 numbers under 400.
The number of divisors of each of the above 19 numbers, with the exception of 1, is the exponent of that number + 1. Example: 2^8 has 8 + 1 = 9 divisors, because 2^8 = 256 which has:1 | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 (9 divisors).....and so on. The ones with more than one factor, such as:18^2 =324 =2^2 x 3^4 =(2+1) x (4+1)=3 x 5 = 15 divisors.....and so on.

idk !

deleted

The possible ways to toss coin B:                 The possible ways to toss coin A:

  H H                                                                          H H H                             H T T

  H T                                                                           H H T                             T H T

  T H                                                                           H T H                             T T H

  T T                                                                           T H H                              T T T

The probability of getting two Heads with coin B and three Heads with coin A:            (1/4)(1/8)  =  1/32.

The probability of getting one Head with coin B and two or three Heads with coin A:  (2/4)(4/8)  =  8/32.

The probability of getting no Heads with coin B and at least one Head with coin A:     (1/4)(7/8)  =  7/32.

Adding together these possibilities:  1/32  +  8/32  +  7/32  =  16/32  =  1/2.

Actually it's 1/2 because you have to subtract the possibilities where it is an equal amount of heads.  That was specified in the answer, so...\(\color{gold}{\colorbox{blue}{you're wrong.}}\)

That's correct.