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J = 84

w = 91

Here's the solution posted by Heureka  :

https://web2.0calc.com/questions/circle-is-the-incircle-of-triangle-abc-and-is-also_1

6 * 87   =   522 points

5 * 88   =  440 points

The grade dropped  is    522  - 440  =    82

A : B : C = 6 : 5 : 3

(area of Pentagon)/(area of Triangle) = 6/3 = 2 = A/C

shaded = 1/4c

unshaded = area of Pentagon - shaded

              = 2c - 1/4c

              = (8c - c)/4

unshaded = 7/4c = 7/4 * A/2

unshaded = 7/8 (area of Pentagon).

unshaded = 7/8 (area of figure)

                     7/8

Call the total number of  marbles in the  bag before we added any  N

The blue  ones=  (4/7)N

The yellow =  (3/7)N

So

[ (4/7)N + 5 ]          13

__________  =   ____    cros-multiply

[ (3/7)N - 3 ]            3

3 [ (4/7)N + 5 ]  =  13 [ (3/7)N - 3 ]    

(12/7)N + 15  =  (39/7)N - 39

(39/7)N - (12/7)N  = 15 + 39

( 27/7)N =  54

N =  54 ( 7/27)  =  14

The number of blue originally =  (4/7)(14)  =   8

Surface area of prism  =

2 [ 23* 84 + 84*28 + 28*23  ] =    9856 in^2

No. of inches of cloth necessary =  ceiling [ 9856/60  ]  ≈  165 in

No. of feet  of cloth  =   165  /12   ≈  13.75 ft

It was 120, but thanks for trying to help

Angle AYX = 135 degrees.

500 grams =  1/2 kilograms

4 (1/2 kg)  =    

2  kg

The area of the parrelogram is \(EG \times DC = 16 \times 32 = \color{brown}\boxed{512}\)

j =  4(5)^2 - 2(2)^2 + 5  =  97

2k  = 2 [  3(1)^2  - 2(4)^2  + 1 ] =   2 [ -28 ]  =   -56

j + 2k  = 

97  -  56   = 

41

Assuming  that A and B could be the same digits  we have

[ 96  - 00  ]  /  4   +   1   =

96 / 4   +   1   =

24  +  1  =  

25 such numbers

Yep.....you are correct, BB  !!!!

It's not possible... 

if the year is odd, then it must end in a 1, 3, 5, 7, or 9.

But, recall that we already have a 2 as the leading digit, so the only way it will work is if it's 2001, but that before 2010, not after 2010. 

The coefficient of x in (x^4 + x^3 + x^2 + x + 1)^5 is 5.

  !

What about \(\color{brown}\boxed{32,222,222}\)?

We have an infinite series, and the formula for the sum is \({ \text{starting term} \over {1 - \text{common ratio}}}\)

Here, the starting term is \({1 \over 7}\) and the common ratio is \(1 \over 7\). 

Can you take it from  here?

333332

[ ( 10!) / (5 * 10) ] mod 17    =     3   

2x^2  = -2y^2 + 12x - 4y  + 28     rearrange  as

2x^2 - 12x + 2y^2 + 4y   =  28          divide through by 2

x^2  - 6x  + y^2  + 2y  =  14             complete the square on x, y

x^2 -6x + 9   + y^2  + 2y + 1  = 14 + 9 + 1

( x - 3)^2  +  ( y + 1)^2  =   24

The radius of the square =   sqrt (24)

The diameter= the side of the  square   =  2sqrt (24)

So.....the area of the square  =   ( 2sqrt (24))^2  =   4 * 24  =   96 units^2

0

To see why....note that    5!  ends with a 0......so.....multiplying this  by any other non-zero units digit(s)  will  always produce a "0"

Mid-point rule

(4 + x) / 2   =  8                 (3 + y) / 2   =  19                   

4 + x  =  8 * 2                      3 + y  =  19 * 2

4 + x  =16                            3+ y = 38

x = 16 - 4                              y =  38 - 3

x = 12                                   y = 35

Sum  =  35 + 12  = 

47

We have 5 choices  for  each  digit....so....

(5) (5)  = 

25

sqrt (1,000,000) -(1,000,000) ^(1/3)  =

1000 -   100  =

900

Each side will be  10 / sqrt (2)  cm long

The area  =  (side)^2  =    (10 / sqrt (2))^2  =      100 / 2 =      50 cm^2