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I have a bad habit of not recognizing tricks, and usually brute force everything... Only when I review do I realize the clever tricks to problems. This is why I take so long on AMC competitions.

The worst kind of problems are the ones where you have to know a formula to solve, because there is no cleverness and you know it or you dont know it.

I mean the trick is to find all unique pairs that added up to 10. But its KIND OF a trick.

Thanks, CU  and Tom  !!!

Brute force! 

Oh no I was hoping for a trick... :(

But your answers are still great!

These problems are really brute force, there is no real clever way. I only just listed all possible combinations

HOWEVER, notice in CPhill's answer.

5 19

5 2 8

5 3 7

5 4 6

Since ALL the sets contain a 5, we just subtract 5 from 15.

So we need TWO terms in a set and they add up to 10, and also are unique.

This is quite easy.

1 9 

2 8 

3 7

4 6

5 5 (doesnt work because cannot repeat digit)

I get four sets

{5  1   9}

{5  2   8}

{5  3   7}

{5  4   6}

{5, 5, 5}  not possible

(5, 6, 4} repeat set

(5, 7, 3}  repeat set

(5, 8, 2} repeat set

(5, 9 , 1} repeat set

When you multiply two numbers, the number of significant figures is the least number of decimal places. In this case, that is 2 (3.15 has 2 decimal places). Rounding 6.319 to two decimal places, you get 6.32

Hope that clarified things

\(\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right) \)

\(\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right) \color{blue} \approx 2.64575131108 \)

  !

48 over 40 divided by 10 over x divided by 4

\(\frac{48}{40}:\frac{10}{x}:4=\frac{48x}{40\cdot 10\cdot4}=\color{blue}0.03x\)

  !

BTW...I'm not an expert on sig figs....but.....here's a site that might help :

https://people.uwec.edu/carneymj/Adobe%20handouts/Significant%20Figures.pdf

In multiplication...the number of sig figs in the final calculated value will be the same as that of the quantity with the fewest number of sig figs used in the calculation.

70.50   = 4 sig figs

10.3   =  3 sig figs

3.00  =  3 sig figs

F  =  mv^2 / r     =    [70.50 kg ] * [3.00 m/s]^2  / [10.3m]   =  61.6  N

3.15  =  3 sig figs

2.006  = 4 sig figs

3.15  * 2.006  = 6.3189  =  6.32 m^2

why is that?

The answer would be C. 6.32 m^2

Hope this Helps ;P

a=2; c=2*log(a) - logn(a, (1/100));printc, a; a++;if(a<100, goto1, 0):
The minimum value occurs at x = 10
2 \log_{10} x - \log_x \frac{1}{100} = 4 - which is the minimum value

The trick is to try to get the bases the same

Note  64  =  4^3

So   we have

(4^3)^(x + 7)  =  (4)^(5x - 3)

4^(3x + 21)  = (4)^(5x - 3)       we  have the bases alike.....solve for the exponents

3x + 21  = 5x  - 3    

24  =  2x

x  = 12

2x  + 2√ [ x ( 1-x) ]  =

2x  + 2 [ x - x^2}^(1/2)       take the derivative and set to 0

2  + 2(1/2) [ x-x^2]^(-1/2) ( 1-2x)  = 0

2 +  (1-2x)

      ____________   = 0

      √  (x-x^2)

2 √ [x- x^2 ]  + 1 - 2x  = 0

2 √ [ x - x^2] =  2x -1    square both sides

4 [x - x^2] =  4x^2 - 4x + 1

4x - 4x^2  = 4x^2 - 4x + 1

8x^2 -8x + 1  = 0

Using the quadratic Formula

x  =  8 ±  √ [ 64 - 32]                8  ±  √ 32               8  ±  4√2         2  ±  √2

      ______________   =      __________  =        ________ =   _______   ≈    .853  or  .146   

                16                               16                          16                    4

When  x = 2 - √2  ,   2x  + 2√ [ x ( 1-x) ]   =    1 

                 _____

                     4

When x  =  2  + √2

                 _______  ,     2x  + 2√ [ x ( 1-x) ]   =    1  + √2  ....and this is the max

                      4

I got one more question:

How do I solve exponetial equations?

For example:

64^x+7=4^5x-3

{We can ignore the subscript 10.......and write }

3*log2 + log(x-4)          {  a log b   =   log b ^a  }

log 2^3  + log (x - 4)       ( log a + log b  =   log (a * b)   }

log 8 + log (x - 4)   

  log ( 8 * (x - 4) )  =

log ( 8x - 32 )

Nvm I fot the correct answer now

Thx for the help but question 2 is incorrect, Thx for quesiton 1 tho 

-wolf

At the end of October of the first year  she will have   (100 - 10)*10  = 90 * 10 =   $900 in her account

In November....she  contributes $100 and she will have  $1000 in her acct at the end of November

In December, she contributes $100  and she will have $1000 + $100  = $1100 in her acct at the end of the year

She  will then make 1% interest on this  so she will have an ending balance of 1100 (1.01)  = $1111   in her account

The second year  she adds  $1200  to this  = $2311  and she makes 1% interest on this  = 2311 (1.01)  =

$2334 at the end of  the second year

And in the third year...she adds $1200 more to this  = 3534 and she makes 1% interest on this and ends up with

3534(1.01) = $3569

CORRECTED ANSWER.......!!!!

\(\alpha+\beta \in \mathbb{R} \Rightarrow Im(\alpha)+2=0 \Rightarrow Im(\alpha)=-2\\ \alpha-2\beta \in i \cdot \mathbb{R}\Rightarrow Re(\alpha) - 6 = 0 \Rightarrow Re(\alpha)=6\\ \alpha = 6-2i\)

1)   11^(5y)   =  (1 /121)^(2y+3)          {Note : 1/121   =  1/11^2  = 11^(-2)   }

So we have

11^(5y)   =  (11^(-2) )^(2y + 3)

11^(5y)  =  (11)^(-4y - 6)        we have the bases the same......solve for the exponents

5y = -4y - 6      add 4y to both sides

9y  = -6              divide both sides by 9

y =  -6/9   =  -2/3

The lateral area  is given by :   pi * R * L      where R is the cone's radius  and L is the slant height

The slant height ,L,  is given by   √[ cone height^2 +  radius^2]  =  √[4^3 + 3^2]= √25  = 5

So....the lateral area  = 

pi * 3 * 5   =

15 pi units^2

why was this question not answred or let alone blocked?