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Thank you Melody!,,2 lessons learnt!!...stay blessed!

Melody is right! You should not use that word on yourself.

You should let us use it on you, because we are professionals and we know how to use it correctly.

Juriemagic, you are definitely not stupid and it is not a word you should use, especially not on yourself :)

All I have said is   

\(10^m\\=10^{m-1+1}\\=10^{m-1}\cdot10^1\\=10^{m-1}\cdot10\\=10\cdot 10^{m-1}\)

If you still can't work it out then say so :)

I am having trouble seeing why ...  ?

\(\frac{DC}{AD}\cdot\frac{AD}{AB}=sina\cdot cos b\)

A geometric series

\(b_1+b_2+b_3+\cdots+b_{10} \) has a sum of 180.
Assuming that the common ratio of that series is \(\dfrac{7}{4}\),
find the sum of the seres \(b_2+b_4+b_6+b_8+b_{10}\).

Geometric series:

\(\begin{array}{|rcll|} \hline b_1 &=& a \\ b_2 &=& ar \\ b_3 &=& ar^2 \\ b_4 &=& ar^3 \\ b_5 &=& ar^4 \\ b_6 &=& ar^5 \\ b_7 &=& ar^6 \\ b_8 &=& ar^7 \\ b_9 &=& ar^8 \\ b_{10} &=& ar^9 \\ \mathbf{r} &=& \mathbf{ \dfrac{7}{4}} \\\\ s_{10} & = & b_1+b_2+b_3+b_4+b_5+b_6+b_7+b_8+b_9+b_{10} \\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{180} \\ \hline \end{array}\)

The sum of a geometric series \(s_{10}\):

\(\begin{array}{|rcll|} \hline s_{10} &=& a\left(\dfrac{1-r^{10}}{1-r}\right) \\ &\text{or} \\ \mathbf{a} & \mathbf{=}& \mathbf{ \dfrac{s_{10}(1-r) }{1-r^{10}} } \\ \hline \end{array} \)

\(\text{Let $\mathbf{x}=b_2+b_4+b_6+b_8+b_{10}$}\)

\(\begin{array}{|rcll|} \hline s_{10} &=& b_1+b_2+b_3+b_4+b_5+b_6+b_7+b_8+b_9+b_{10} \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+b_3+b_5+b_7+b_9 \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\left(\dfrac{b_3}{r}+\dfrac{b_5}{r}+\dfrac{b_7}{r}+\dfrac{b_9}{r}\right) \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8) \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8)+rb_{10}-rb_{10} \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8+b_{10})-rb_{10} \\\\ s_{10} &=& b_1+x+r\cdot x-rb_{10} \\\\ s_{10} &=& x(1+r) + b_1-rb_{10} \quad | \quad b_1 = a,\ b_{10}= ar^9 \\\\ s_{10} &=& x(1+r) + a-rar^9 \\\\ s_{10} &=& x(1+r) + a-ar^{10} \\\\ s_{10}&=& x(1+r) + a(1-r^{10}) \quad | \quad \mathbf{a = \dfrac{s_{10}(1-r) }{(1-r^{10})} } \\\\ s_{10}&=& x(1+r) + \dfrac{s_{10}(1-r) }{(1-r^{10})}(1-r^{10}) \\\\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{x(1+r) + s_{10}(1-r) } \\ \hline \end{array} \)

\(\mathbf{x=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_{10}} &\mathbf{=}& \mathbf{x(1+r) + s_{10}(1-r) } \\\\ x(1+r) &=& s_{10} - s_{10}(1-r) \\ x(1+r) &=& s_{10}\Big(1 - (1-r) \Big) \\ x(1+r) &=& s_{10} (1 - 1+r ) \\ x(1+r) &=& s_{10}r \\ \mathbf{x} &\mathbf{=}& \mathbf{s_{10}\left(\dfrac{r}{1+r}\right)} \quad | \quad s_{10}=180,\ r=\dfrac{7}{4} \\ x & = & 180\cdot \left(\dfrac{\dfrac{7}{4}}{1+\dfrac{7}{4}}\right) \\\\ x & = & 180\cdot \left(\dfrac{\dfrac{7}{4}}{ \dfrac{11}{4}}\right) \\\\ x & = & 180\cdot \left( \dfrac{7}{11} \right) \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{1260}{11}} \\ \hline \end{array}\)

The sum of the series \(\mathbf{b_2+b_4+b_6+b_8+b_{10}}\) is \(\mathbf{\dfrac{1260}{11}}\)

thank you Melody,

please forgive my stupidness...I have now studied your second step....I don't understand how we can just multiply the bottom part by 10. which math law is at work here?...from step 3 onwards I fully understand, it really is just step 2...would you mind teaching me please?

I'm always pleased to help you :)

Melody!!!!!..always to the rescue..lol....thank you!!

You know..this must be telepathy of some sort because I also came across a different approach for this sum in which there was this piece:

\((10-2)*10^{m-1}\)

and I was about to ask here for an explanation as to how this step is achieved, and walla!!..you have it in your response!!..thank you again Melody!

Yes i understand that.    

Sometimes I think mathematicians (not you) purposely design things to be complicated when it does not seem remotely necessary. 

Hi Somebody :)

Where did you get CPhill's answer from?

It is usually nicer to give the link.   (I gave you a point though) Thanks for answering   :)

Lets see

\(\frac{53^{2000}}{53^{1999}}=53^{2000-1999}=53\)

Think about the second one.

The numerator and the denominator are reduced by 2000

    but proportionately the numerator has been reduced by less than the denominator because it was bigger to start with.

   so proportionately... the top was reduced by a smaller factor than the bottom so the entire number got bigger.

---------------------

It is like this

\(\frac{10}{8}\;\;compared \;\;to \;\;\frac{9}{7}\\ 1\frac{2}{8}\;\;compared \;\;to \;\;1\frac{2}{7}\\ \)

The second one is bigger.

The sum of a geometric series whose first three terms are 8000, -12000, and 18000 is 57875.

What is the last term of the series?

Geometric series:

\(\begin{array}{lrcll} \text{first term} & a_1 &=& a \\ \text{second term} & a_2 &=& ar \\ \text{third term} & a_3 &=& ar^2 \\ \ldots \\ \text{last term} & a_n &=& ar^{n-1} \\\\ \text{so}& a_1 &=& 8000 \\ & a_2 &=& -12000 \\ & a_3 &=& 18000 \\\\ \text{a=?} & a &=& a_1 \\ & \mathbf{a} & \mathbf{=}& \mathbf{8000} \\\\ \text{r=?} & r &=& \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} \\ & r &=& \dfrac{-12000}{8000} \\ & \mathbf{r} & \mathbf{=}& \mathbf{-\dfrac{3}{2}} \\ \end{array} \)

The sum of a geometric series \(s_n\):

\(\begin{array}{|rcll|} \hline s_n &=& a\left(\dfrac{1-r^{n}}{1-r}\right) \\\\ s_n &=& \dfrac{a-ar^{n}}{1-r} \\\\ s_n &=& \dfrac{a-ar^{n-1}r}{1-r} \quad | \quad a_n = ar^{n-1} \\\\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{a-a_nr}{1-r} } \\ \hline \end{array}\)

The last term \(a_n\):

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{a-a_nr}{1-r} \\ s_n(1-r) &=& a-a_nr \\ a_nr &=& a-s_n(1-r) \\ a_n &=& \dfrac{ a-s_n(1-r) } {r} \quad & | \quad a=8000,\ s_n =57875, \ r =-\dfrac{3}{2} \\ a_n &=& \dfrac{ 8000-57875\left(1-\left(-\dfrac{3}{2}\right)\right) } {-\dfrac{3}{2}} \\ a_n &=& -\dfrac{2}{3}\cdot \left( 8000-57875\cdot \dfrac{5}{2} \right) \\ a_n &=& \dfrac{1}{3}\cdot \left( 5\cdot 57875-16000 \right) \\ a_n &=& \dfrac{273375}{3} \\ \mathbf{a_n} & \mathbf{=}& \mathbf{91125} \\ \hline \end{array}\)

The last term of the series is 91125

Thanks itsyaboi,

I'd just like to have a go at it too :)

\({5^{m-2}*2^{m+2}} \over 10^m-10^{m-1}*2\\~\\ =[{5^{m-2}*2^{m+2}} ]\div [10^m-10^{m-1}*2]\\~\\ =[{5^{m-2}*2^{m-2}*2^4} ]\div [10*10^{m-1}-10^{m-1}*2]\\~\\ =[10^{m-2}*2^4 ]\div [10^{m-1}(10-2)]\\~\\ =[10^{m-2}*16 ]\div [10^{m-1}(8)]\\~\\ =10^{m-2-(m-1)}*2 \\~\\ =10^{-1}*2 \\~\\ =\frac{2}{10}\\~\\ =\frac{1}{5}\)

goodness me!!,

Rom, every year the math is becoming more and more complicated...this is for a grade 11 pupil...I cannot believe the complexity of the sums nowadays...thank you very much Rom, I definitely would never have been able!..thank you!!

wow!,

thank you sooo much!..really appreciated!!

\(2^{x+3}-3\cdot 2^{x-1} = 104\\ 8\cdot 2^x - \dfrac 3 2 2^x = 104\\ \dfrac{13}{2}\cdot 2^x = 104 \\ 2^x = 16\\ x = 4\)

The second one is a tad trickier

\(2^{2x} + 7 \cdot 2^x = 8\\ u=2^x\\ u^2 + 7u - 8 = 0\\ (u+8)(u-1) = 0\\ u = -8, 1\\ x = \log_2(u)\\ \text{-8 is seen to be an extraneous solution}\\ x = \log_2(1)=0\\ \text{and indeed}\\ 2^0 + 7 \cdot 2^0 = 1+7=8\)

Factoring the bottom yields (1-1/5) times 10^m

Now, you have the fraction: (5^(m-2 * 2^(m+2)))/((4/5)*10^m)

10^m can be expressed as 2^m * 5^m

Cancelling yields:

4/(5^2 times 4/5)

4/20 = 1/5

well he did ask to use the formal definition of a limit  

Thank you, CPhill

Solution (partial):

\(\text {The equation for combined gas laws: }\\ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\\ \text { }\\ \text {The solution requires simple algebraic operations for this setup: }\\ \text { }\\ \dfrac{1.00 atm*2.00 L}{T_1} = \dfrac{0.82atm*2.00L}{10 C} \quad | \quad \text {solve for } T_1\\ \text { }\\ \text {Algebra 1 & 2 is prerequisite for chemistry; you should already have these skills. }\\ \text { }\\ \text {Post your (attempted) solution below. Someone will check it for you. }\\ \)

GA

2) b + bx +bx^2 +bx^3 .... +bx^9 = 180, where x =7/4

Therefore, b= 47185920/93808891

Therefore, b_2+b_4+b_6+b_8+b_10 = 1260/11.

1) By the geometric series (finite) formula,

 \(\frac{8000(1-(-1.5)^n)}{2.5}=57875 \\ \text{where n is the number of terms}\)

n=7, so 8000 times -1.5^7 = -136687.5.

center of the squares

2. CPhill's Answer:

Let one of the integers   = a

Let the other integer  = b

And 1 less than a multiple of 5 can be written as  5n  - 1   where n is an integer ≥ 1

So....we have this equation

  S          +   P   =  5n - 1

(a + b )   +  (ab)  = 5n - 1      (1)

Rearranging (1), we have

a + ab +  b  = 5n -1

a + ab + b + 1  = 5n

a (b + 1)  + 1 ( b + 1)   = 5n

(a + 1)(b + 1)  = 5n

Note that if  "a" ends in a "4"  or a "9' then  (a + 1)  is a multiple of 5, and no matter the integer value of b, the left side is always a multiple of 5. And the right side is definitely a multiple of 5

So...the  "a's"  that  end in either 4 or  9 from 1-50 inclusive are :

4, 9 ,14, 19, 24, 29,34, 39, 44, 49

Notice that we can pair 4 with any of the other 49 integers and (1) will be true

Likewise, we can can pair 9 with any of 48 integers  [we've already paired it with 4 ], and (1)  will be true

And 19 can be paired with any of 47 other integers  [ we've already paired it with 4 and 9 ] and (1)  will be true

Continuing this reasoning with each successive number, we finally arrive at the fact that 49 can  be paired with any of 40 other integers and (1) will be true

So....the number of pairs  is just   49 + 48  + 47 +  .....+ 41  + 40

And totalling these, we have that the number of pairs is just :

( 49 + 40)  *  10  / 2   =

89 * 5   =

445  pairs a,b   that make (1)  true

And the number of possible pairs of a and b is  C(50,2)   = 1225

So, the probability that P + S  is one less than a multiple of 5, is just

445  / 1225  =

89  / 245    ≈  36.3 %

Good spotting CU.

Clelz, No one can read that, or at least I cannot anyway. You need to post your question on the actual forum.