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Note that

1111111=10^6+10^5+10^4+10^3+10^2+10+1

and

909091=10^6-10^5+10^4-10^3+10^2-10+1.

Compute the product of these two integers.

geometric series:

\(\begin{array}{|rcll|} \hline a_n &=& a_1 \cdot r^{n-1} \\ s_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \qquad \text{sum}\\ \hline \end{array}\)

geometric series 1:

\(\begin{array}{|rcll|} \hline && 10^6+10^5+10^4+10^3+10^2+10+1 \qquad | \qquad a_1 = 1 \qquad r = 10 \\\\ s_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \\ s_n &=& 1\cdot \left( \frac{-1+10^n}{1-10} \right) \\ s_7 &=& 1\cdot \left( \frac{-1+10^7}{1-10} \right) \\ s_7 &=&\left( \frac{-1+10^7}{1-10} \right) \\ \hline \end{array} \)

geometric series 2:

\(\begin{array}{|rcll|} \hline && 10^6-10^5+10^4-10^3+10^2-10+1 \qquad | \qquad a_1 = 1 \qquad r = -10 \\\\ S_n &=& a_1 \left( \frac{-1+r^n}{1-r} \right) \\ S_n &=& 1\cdot \left( \frac{-1+(-10)^n}{1-(-10)} \right) \\ S_7 &=& 1\cdot \left( \frac{-1+(-10)^7}{1+10} \right) \\ S_7 &=&\left( \frac{-1-10^7}{1+10} \right) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && 1111111 \cdot 909091 \\ &=& (10^6+10^5+10^4+10^3+10^2+10+1) \cdot (10^6-10^5+10^4-10^3+10^2-10+1) \\ &=& s_7 \cdot S_7 \\ &=& \left( \frac{-1+10^7}{1-10} \right) \cdot \left( \frac{-1-10^7}{1+10} \right) \\ &=& \left[ \frac{(-1)^2-(10^7)^2}{1^2-(10)^2} \right] \\ &=& \frac{1-10^{14}}{1-10^2} \\ &=& \frac{10^{14}-1}{10^2-1} \\ &=& \frac{99999999999999}{99} \\ &=& 10101010101010 \\ \hline \end{array} \)

Most likely they do love you.  A lot of teenagers feel like this but it is rarely true.

1111111=10^6+10^5+10^4+10^3+10^2+10+1

and

909091=10^6-10^5+10^4-10^3+10^2-10+1.

Compute the product of these two integers.

This is one way - I did it as the difference of 2 squares.

Heureka is doing it now as well, he probably has a better way...

\((10^6+10^5+10^4+10^3+10^2+10+1)\times (10^6-10^5+10^4-10^3+10^2-10+1)\\ = (10^6+10^4+10^2+1+10^5+10^3+10)\times (10^6+10^4+10^2+1-10^5-10^3-10)\\ =[ (10^6+10^4+10^2+1)+(10^5+10^3+10)]\times[ (10^6+10^4+10^2+1)-(10^5+10^3+10)]\\ =[ (10^6+10^4+10^2+1)^2-(10^5+10^3+10)^2]\\ =[ (10^6 (10^6+10^4+10^2+1)+ 10^4(10^6+10^4+10^2+1)+10^2 (10^6+10^4+10^2+1)+(10^6+10^4+10^2+1)] -(10^5+10^3+10)^2]\\ =[ 10^{12}+2*10^{10}+3*10^8+4*10^6+3*10^4+2*10^2+1]-[(10^5(10^5+10^3+10)+10^3(10^5+10^3+10)+10(10^5+10^3+10)]\\ =[ 10^{12}+2*10^{10}+3*10^8+4*10^6+3*10^4+2*10^2+1]\\ \qquad \quad-[10^{10}+2*10^8+3*10^6+2*10^4+10^2]\\ =[ 10^{12}+10^{10}+10^8+10^6+10^4+10^2+1]\\ =1010101010101 \)

-14/24

The answer is 378,000. Just subtract 728,000 from 1,106,000

This answer is -2/5.

p=v x .89^t - Annual decline in house prices.

p=v x 0.990336^(12t) - Monthly decline in house prices

1:2

Changing them to mod 10

m=9a+6=3(3a+2)

n=9b=3×3b

a and b are integers.

It is possible that n is 0 all (any mod) but if it is not the answer is 3.

1111111*909091=1,010,101,010,101= 239 * 4649 * 909091

Evaluate 3/2+5/4+9/8+17/16+33/32+65/64-7.

= - 1/64

lol

16/1997 is the 5th convergent in it's continued fraction expansion.

what is 2 and five eights minus 1 and one third,

2 5/8 - 1 1/3 =

21/8 - 4/3 = The GCD of 3 and 8 is 24

63/24 - 32/24 =

[63 - 32] / 24 =

31 / 24 =1 7/24

Just use a calculator. To explain to you in detail will give you a headache!!!.

75 3/4 - 1/2=75.75 - 0.50=75.25 =75 1/4

First enter your number and then press "+-" key just to the right of zero. That will change the sign of your number.

The antilog is given by  :

10^[0.119280313 ]  ≈  1.316

a) How long will it take for her money to triple?

Let A be the beginning amount .....note 5+3/8% = .05375....  and we have

3A  = Ae^(.05375 t)     divide both sides by A

3  = e^(.05375t)    tkae the ln   of both sides

ln 3  = ln e ^ (.05375t)   and we can write

ln 3   = (.05375t) ln e       and since  ln e  = 1, we can ignore this

ln 3 = (.05373t)     divide both sides by  .05735

ln (3) / .05375   = t =  about  20.44 years

b) How long will it take for her money to sextuple?

The process is similar to the first one......at the end we will have

ln (6)  / .05375   =  t = about   33.34  years

CPhill: I have programmed my computer (in Java) to convert decimals to fractions and vise versa, up to about 50 digits long. It does it instantaneously.

Could you explain how you arrived at this solution???

\(25=\frac{24^{2}}{a^{2}}+\frac{a^{2}}{2^{2}}\)

25  = 576/a^2  + a^2 / 4        getting a commom denominator on the right, we have

25  = [ 576 * 4 + a^4 ]  / [ 4a^2]    cross-multiply

25 * 4a^2  = 576 *4 + a^4

100a^2  = 2304 + a^4    rearrange

a^4 - 100a^2 - 2304  =  0      this can be factored as

(a^2 - 36) (a^2 - 64)   = 0      setting both factors to  0, we have that

a^2 - 36  = 0        which means that a = ± 6    and

a^2 - 64  = 0       which means that the other two soliutions are a = ± 8

So  a =  { -8, -6, 6, 8 }

a) It will Victoria 20.44 years to triple her money @ 5 3/8% comp. continuously.

b) It will Victoria 33.34 years to sextuple her money @ 5 3/8% comp. continuously.

f(x)   = log₂ (4x)  - 2       write f(x)  as y

y =   log₂ (4x)  - 2        add 2 to both sides

y + 2   = log₂(4x)  

This says that, in exponential form :

2^{y + 2}   = 4x           divide both sides by 4

[2^{y + 2} ] / 4   = x       now, exchange x and y

[2^{x + 2} ] / 4  = y       and for y, we can write  f^-1(x)  = g(x)

[2^{x + 2} ] / 4  = g(x)     and this is the inverse

3^(81x) = 81^(3x)

Let's do this by equating bases

Note   81 =  3^4       so we have

3^(81x)   = [3^4]^(3x)     simplify

3^(81x)   = 3^(12x)      since we have the bases the same, we can solve for the exponential equations

81x  = 12x    

81x - 12x = 0

(81 - 12)x = 0 

69x = 0       and the only thing that makes this true is when x = 0