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Frank answered 76 questions correctly.

Draw a tangent between two parabolas.  This leads to an answer of 5/8.

The domain of g(x) is [-1,1], so its width is 2.

f(g(1)) + g(f(1)) = 14.

The probability is 4/1009.

Here's a hint: draw the other half of the semicircle, and drop perpendiculars.

a - b = 144*pi - 16*pi = 128*pi.

The probability is 3*C(7,3)/C(21,4) = 1/57.

Just use the quadratic formula!  It's easy.

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The first term of an arithmetic progression is -8. The ratio of the 7th term to the 9th term is 5:8. Calculate the common difference of the progression  

–8      –5      –2      1      4      7      10      13      16  

1st     2nd    3rd                             7th               9th

The common difference is +3  

_.

Let George's age be x. Then Alex's age is x+12 and Carl's is 3(x+12). Their ages add up to 68 so we can form an equation.

x+(x+12)+3(x+12)=68

2x+12+3x+36=68

5x+48=68

5x=20

x=4

So George's age is 4, Alex's is 16, and Carl's is 48. The ratio of their ages is:

4:48:16

=1:12:4

I just finished my week 12 AoPS prealgebra 2149

what is x=4 in standard form. ax^2 +bx +c

Hello Guest!

\((x-4)(a-4)=(x-4)^2=x^2-8x+16\)

\(ax^2 +bx +c\)

\(a=1\\ b=-8\\ c=16\\ \)

\(x^2-8x+16\)

  !

Allen finishes in 12 seconds

    Bruce still has 3 seconds to go to reach the finish his 15 second run

          so Bruce still has  3/15 of the race to go    3/15 x 100 = 20 m left to run

rise / run = 3% = .03

x/ 1800 = .03

x = 54 m

Each of the diagonals is an arithmetic series.

The formula for the sum of an arithmetic series is:  Sum  =  n(a₁ + a_n) / 2

where         n = number of terms         a₁ = first term         a_n = last term

For the diagonal from upper left to lower right:  n = 100     a₁ = 1     a_n = 10000

--->     Sum  =  100( 1 + 10000 ) / 2  =  500 050

For the diagonal from upper right to lower left:  n = 100     a₁ = 100     a_n = 9901

--->     Sum  =  100( 100 + 9901 ) / 2  =  500 050

You'll need to add these two sums together to get the final answer ...

COmmon ratio    r      1st term 6    multiplied by common ratio yields second term

                                        so   r = -1

this sequence just bounces back and forth between  6    and  -6  

    even numbered terms are negative  (like the second term above)

       odd numbered terms are positive

205 th term would be   +6

The usual amplitude of a sin or cos wave is  1     and they go from +1 to -1   ( a height of 2   amplitude = height/2)

   this wave goes from -14 to -2   a height of 12      height/2 = 6 for amplitude

so 'a' in your equation is 6   it multiplies the wave by 6

6 cos (bx+c ) + d

'd' shifts the wave up or down.... the usual sin or cos wave has a midline at  y = 0

   ths one has a midline at y = -8    shifted down 8

6 cos (bx+c) - 8

'b' stretches or compresses the wave

     usual period for sin or cos wave is 2pi

        this one has 1/2 of a period from  3/4 pi to 2 pi     full period is   2* (2pi- 3/4 pi ) = 10/4 pi

           it is stretched out by 'b'       2pi / ( 10/4 pi) = 8/10 = 4/5 =b

6 cos (4/5 x +c) -8    

      'c'  shifts the wave left or right ...

           this cos wave is shited RIGHT  3/4 pi     (the high point of cos wave is usually at 0 this one is at 3/4 pi)

                SUBTRACTING SHIFTS RIGHT

                     BUT wecompressed the wave with 'b'  so 'c' will need to be compressed too

                           3/4 pi  * 4/5 = 3/5 pi

6 cos (4/5 x -3/5 pi ) - 8  

Thank you!!

The shaded area is everything ABOVE and INCLUDING the line

    so look for    y>=       Only one choice fits this criteria

Thank you very much!

Since you only have 3 numbers, that means the entire number of palindromes will begin with one of the 3 numbers. And since the 3 numbers can be repeated, then each sequence of palindromes beginning with any of the 3 numbers will have:3^3 palindromes =27. And since you have only 3 numbers, then the total number of palindromes will = 27 x 3 = 81 palindromes. So, if you count the number of palindromes beginning with 7, 8 and 9, you should get 27 for each of the 3 numbers for a total of 81 such numbers.

Thanks :D

However, is there an easier way than counting everything?