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It's not A and C...

Find angle BPC     

angle PBC = 150^o which is the total of PBA (60^o cuz the triangle is equilateral) plus ABC (90^o cuz it says so)  

since PBC is 150^o that leaves 30^o for the total of both BPC and BCP cuz the total of angles in a triangle is 180^o  

triangle PBC is isoceles since sides PB and BC are equal  

therefore angles BPC and BCP are equal, so each one is half of their 30^o total  

so angle BPC is 15^o cuz 15 is half of 30

_.

The total area of the pieces is 4*5 = 20, so the smallest possible area of the rectangle is 20.

You can't square a real number and obtain a negative product.  

thank you!!

Let's say they have altogether x at first

x/2 + 65 = (x/2 - 23) * 3

 x + 130 = 3x - 6x * 3

 268 = 2x

   x = 134

$\angle ACD = 22^\circ$.

n = r
(n+65) = 3 (r -23)
n + 65 = 3 (n-23)
n + 65 = 3 n - 69
134 = 2 n
n = r = 67
so
n + r = 134

There are 6 terms adding up to 63 / 128 as follows:

1 / 4 , 1 / 8 , 1 / 16 , 1 / 32 , 1 / 64 , 1 / 128 , >>Number of terms = 6>>Total Sum == 63 / 128

Let the number of polished stones that Nora and Renata had =N =R

N == R

R + 65==3 [R - 23]

R + 65==3R - 69

3R - R ==69 + 65

2R ==134

R == 134  polished stones they started with, or 134/2==67 stones each

There are 150 such sequences as follows:

14556 , 14565 , 14655 , 15456 , 15465 , 15546 , 15564 , 15645 , 15654 , 16455 , 16545 , 16554 , 22556 , 22565 , 22655 , 23455 , 23545 , 23554 , 24355 , 24535 , 24553 , 25256 , 25265 , 25345 , 25354 , 25435 , 25453 , 25526 , 25534 , 25543 , 25562 , 25625 , 25652 , 26255 , 26525 , 26552 , 32455 , 32545 , 32554 , 34255 , 34525 , 34552 , 35245 , 35254 , 35425 , 35452 , 35524 , 35542 , 41556 , 41565 , 41655 , 42355 , 42535 , 42553 , 43255 , 43525 , 43552 , 45156 , 45165 , 45235 , 45253 , 45325 , 45352 , 45516 , 45523 , 45532 , 45561 , 45615 , 45651 , 46155 , 46515 , 46551 , 51456 , 51465 , 51546 , 51564 , 51645 , 51654 , 52256 , 52265 , 52345 , 52354 , 52435 , 52453 , 52526 , 52534 , 52543 , 52562 , 52625 , 52652 , 53245 , 53254 , 53425 , 53452 , 53524 , 53542 , 54156 , 54165 , 54235 , 54253 , 54325 , 54352 , 54516 , 54523 , 54532 , 54561 , 54615 , 54651 , 55146 , 55164 , 55226 , 55234 , 55243 , 55262 , 55324 , 55342 , 55416 , 55423 , 55432 , 55461 , 55614 , 55622 , 55641 , 56145 , 56154 , 56225 , 56252 , 56415 , 56451 , 56514 , 56522 , 56541 , 61455 , 61545 , 61554 , 62255 , 62525 , 62552 , 64155 , 64515 , 64551 , 65145 , 65154 , 65225 , 65252 , 65415 , 65451 , 65514 , 65522 , 65541 , Total =  150 such sequences.

The number of students at first was 3/2 times the number of teachers.

      How many teachers were in the school hall?
 

 given, number of students at first

 = 3/2 times the number of teachers

=> number of teachers = 2/3 number of students

     
      = 2/3 * 180 = 2 * 60 = 120

So there were 120 teacher in the school hall

The value of n is 5.

5 -  1 1/4  - 1 1/2  =

20/4    - 5/4    -   6/4   =    (20-5-6)/4 =  9/4 = 2  1/4    lb

(3 3/4) =  30/8

(30/8)  /  (3/8)   = 30/8 * 8/3 = 30/3 = 10 times

.5 x = 35 dollars         x = max book price,   solve for 'x'

EP: Just a minor typo: 0.55 x 36 2/3 ==20 + 1/6

13 out of (13 + 14 +9) =  13 / 36ths

13/36 * 288 = ------------

Modular multiplicative inverse of 27 modulo 29==14

Check: [27 *  14] mod 29 ==1 - which is the definition of modular multiplicative inverse.

Here are five cards. 1 5 7 9 11 One of the cards is removed.

The mean of the numbers on the remaining four cards is 6

Which card was removed     

The sum of all five cards (i.e., 1 + 5 + 7 + 9 + 11) is 33.  

For the average of four of these cards to be 6, we want their total to be 24.  

That's because 24 / 4 = 6.  

To change the total from 33 to 24, we will remove the 9 card.  

_.

1/2   b * h   = 48

1/2   b *  8  = 48

b = 48 / ( 1/2 * 8) ft

How many Pokemon cards does Caleb have now if they have a total of 

    288 Pokemon cards?

c =3   =  9

m=4   =12

s =5   =15

then

c = 9+4          = 13

m = 8 + 6       =14

s= 9               = 9 

13:14:9

https://web2.0calc.com/questions/geometry_84061