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Thanks for answering Jugoslav.

However please don't state that a question is very simple because it sounds really insulting to the question asker.

In this case, I for one, do not think the question is as simple as you suggest.

Fuhsah, 

Jugoslav has given you some insight which may be enough for you.

If it is not enough then please ask for more help.

State what you have done to try and help yourself though.

Thank you for your answers. That's a lot more than I expected. Thank you.

But what happens to the dollars in my trunk?

  !

x=2y-2

This is very simple. 

ΔABE ~ ΔCDE

BE = x         CE = y

x/18 = y/12

12x = 18y

y = 12x / 18

18 / 21-y = 12 / 19-x

18(19-x) = 12(21-y)

18(19-x) = 12(21- 12x/18)

x = 9         y = 6

I'll let you finish up your homework. 

Triangles ABE and CDE are similar.

101 , 102 , 103 , 104 , 105 , 106 , 107 , 108 , 109 , 110 , 120 , 130 , 140 , 150 , 160 , 170 , 180 , 190 , 201 , 202 , 203 , 204 , 205 , 206 , 207 , 208 , 209 , 210 , 220 , 230 , 240 , 250 , 260 , 270 , 280 , 290 , 301 , 302 , 303 , 304 , 305 , 306 , 307 , 308 , 309 , 310 , 320 , 330 , 340 , 350 , 360 , 370 , 380 , 390 , 401 , 402 , 403 , 404 , 405 , 406 , 407 , 408 , 409 , 410 , 420 , 430 , 440 , 450 , 460 , 470 , 480 , 490 , 501 , 502 , 503 , 504 , 505 , 506 , 507 , 508 , 509 , 510 , 520 , 530 , 540 , 550 , 560 , 570 , 580 , 590 , 601 , 602 , 603 , 604 , 605 , 606 , 607 , 608 , 609 , 610 , 620 , 630 , 640 , 650 , 660 , 670 , 680 , 690 , 701 , 702 , 703 , 704 , 705 , 706 , 707 , 708 , 709 , 710 , 720 , 730 , 740 , 750 , 760 , 770 , 780 , 790 , 801 , 802 , 803 , 804 , 805 , 806 , 807 , 808 , 809 , 810 , 820 , 830 , 840 , 850 , 860 , 870 , 880 , 890 , 901 , 902 , 903 , 904 , 905 , 906 , 907 , 908 , 909 , 910 , 920 , 930 , 940 , 950 , 960 , 970 , 980 , 990 , >>Total = 162 such integers

Shaded area = 7.85398163 square units

or

A = 2.5pi     = 5pi/2

These questions are so fun. Let me know if you need helpwith any more

4 is the answer

(58 + 1/4x) (1 - 1/5) = 95

  (58 + 1/4x) * 4/5 = 96

      58 + 1/4x = 120

              1/4x = 62

                   x = 248

area of the shape.

Let the total pens produced in 5 days be x.

Pens produced on the first day = x/5

second day = 28

third day = (x/5 + 28)/2

4th day = x/5 + 9

5th day = 64

∴, x/5 + 28 + (x/5 + 28)/2

          + x/5 + 9 + 64

          = x

=> x/5 + 1/2(x/5) + 1/2(28)

        + x/5 + 73 + 28 = x

=> x/5 + x/10 + x/5 + 101 + 14 = x

=> \( {2x + x + 2x \over 10} + 115 = x\)
=> \( {5x + 1150\over 10} = x\)

=> 5x + 1150 = 10x

=>  5x = 1150

=>    x = 230

∴, total pens = \(\boxed{230}\)

Let,

     a = Peter's sticker

     b = Jordan's sticker

     c = Albert's sticker

Given:

    1) \({a + b \over 2} = 924\) (average number of Peter's & Jordan's stickers)

         a + b = 2(924)

         a + b = 1848 (1)

     2)

         a = b + 232 => a - b = 232 (232)

     3) 

         c = a + 172 (3)

Solve for a using eq. (1) & (2), eliminate b

               a + b = 1848

            + a - b = 232

                    2a = 2080

                    2a/2 = 2080/2

                      a = 1040

Find b.                                    Find c.

        a + b = 1848                        C = a + 172

              b = 1848 - a                       = 1040 + 172

                 = 1848 - 1040             C = 1212

              b = 808

So,

      \( {Jordan’s stickers \over Albert’s stickers} = {b \over c} = {808 \over 1212}\)

                                         = 0.6666. . .

        Change 0.6666... to fraction.

                     x = 0.6666 . . .

                 10x = 6.6666 . . .

                -    x = 0.6666 . . .

                   9x = 6

                     x = 6/9 = 2/3

Therefore,

Jordan's stickers is 2/3 of Albert's stickers.

                    2/3 * 1212 = 808 ✓

or 5:13

Let the total items bought be x.

Erasers = 4x/11, Pens = 7x/11

Erasers = $1.20, Pen = $1.60

                  4x/11 * 1.2 + 7x/11 * 1.6 = 64

                       4.8x + 11.2x = 64 * 11

                                   16x = 64 * 11

                                       x = 4 * 11   =>  x = 44

No. of erasers = (4 * 44)/11, No. of pens = (7 * 44)/11

                        = 16                                  = 28

You can just plot the graphs, and apply the Horizontal Line Test.  The graphs make it obvious.

p = 16.

k = 44.

The sum of all possible values of k is 21.

Let Kate's mother's age when she was born==M
Let her brother's age when she was born ==B
Her brother's age: B =1/4M - 4

8 years later, Kate's mother was M+8==4B + 4 + 8.........(1)
Her brother's age was: 
B+8 ==1/3[M+5+8]..........................(2), solve for M, B

B==7 years - when Kate was born
M==32 years - when Kate was born.

7 + 8==15 years - her brother's age now
32 + 8 ==40 years - her mother's age now.

Subtract functions: f(x) - g(x) = (3x^2 - 2x + 4) - (x^2 - kx - 6) = 2x^2 - (k + 2) x + 10

Plug in x = 10: f(10) - g(10) = 200 - 10(k + 2) + 10 = 10

Solve for k: k = 8

The equation of the cardioid is \(r = 1 - \sin (\theta)\)

The volume of the cone is 42.

The number of minimal paths is \(\displaystyle \binom{4}{1} + \binom{4}{2} + \binom{4}{3} = 14\)

total distance / total time = average speed

d / 50  = time1

d /75  = time 2

total distance = 2d   

2d /(d/50 + d/75)   = avg speed

= 2 / (1/50 + 1/75) =  60 m/hr average