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pi  x  d   =   circumference

or

pi  x  2r = 25 cm

r =  25 cm / (2pi)  = .......  cm      

Let the price of a single vase =V

let the price a single bunch of flowers =F

2V  +  8F = 15........................(1)

20V  + 5F=25........................(2)

Now, you have 2 simultaneous equations. Can you solve them?

(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 72, 75, 80, 90, 96, 100, 108, 120, 135, 144, 150, 160, 180, 200, 216, 225, 240, 270, 288, 300, 360, 400, 432, 450, 480, 540, 600, 675, 720, 800, 864, 900, 1080, 1200, 1350, 1440, 1800, 2160, 2400, 2700, 3600, 4320, 5400, 7200, 10800, 21600) >>Total = 72 such numbers.

a=0;b=0;c=0;d=0;e=0;p=0; cycle:n= a*10000+b*1000+c*100+d*10+e;if(n%7==0 and n%6==4 and n%9==4 and n%15==4 and n%18==4, goto loop, goto next); loop:printn,", ",;p=p+1; next:e++;if(e<10, goto cycle, 0);e=0;d++;if(d<10, goto cycle, 0);e=0;d=0;c++;if(c<10, goto cycle,0);c=0;d=0;e=0;b++;if(b<10, goto cycle,0);e=0;d=0;c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT = 364 - This is the smallest multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15 and 18.

I'm not sure if I understand your question, but here is my best effort. Notice that I have included numbers 1, 2, 3 and 4, being the even result of: 30/30, 60/30, 90/30 and 120/30. If you are not supposed to include those numbers, Just subtract them from the total:

1 / 30, 7 / 30, 11 / 30, 13 / 30, 17 / 30, 19 / 30, 23 / 30, 29 / 30, 1, 31 / 30, 37 / 30, 41 / 30, 43 / 30, 47 / 30, 53 / 30, 59 / 30, 2, 61 / 30, 67 / 30, 71 / 30, 73 / 30, 77 / 30, 79 / 30, 83 / 30, 89 / 30, 3, 91 / 30, 97 / 30, 101 / 30, 103 / 30, 107 / 30, 109 / 30, 113 / 30, 119 / 30, 4, 121 / 30, 127 / 30, 131 / 30, 133 / 30, 137 / 30, 139 / 30, 143 / 30, 149 / 30 =3251 / 30

This is best tackled as follows:

2 + 4 + 6 + 8 + ... + 100 = 2*(1 + 2 + 3 + 4 + ... + 50)

The sum of the first n integers is given by n(n+1)/2 so the above sum becomes 2*50*51/2 = 50*51 = 2550

a = 7  and   b = 5.  12075 is a multiple of 3, 5 and 7.

Zero appears 192 times between 1 and 1000 inclusive.

https://brainly.com/question/5013675

Oh I see...I didn't see the OP's working. However, I don't understand how the equation leading back to h/a=b/(x+y) proves that (x+h)^2+(y+h)^2=(a+b)^2.

I don't know much about sets and subsets! But it looks to me, just by simple inspection, that set A has 8 elements and set B has 6 elements, only in this sense: Set A =2^8 =256 subsets and Set B =2^6 = 64 subsets.

So: 256 - 64 =192 more subsets in Set A than Set B. Somebody should correct me if I'm wrong.

Will need to know how big the soup labels are  (cm²)

5 students solved neither question  leaving 55

35 + 28 = 63      which is 8 more than 55   ....so there must be an '8' overlap .....

how did we know the "8" between a & b?

There are 400 integers between 1 and 1000 that are neither multiples of 2 nor multiples of 5. Basically ALL the ODD numbers except the ones that end in 5: [1, 3, 7, 9, 11......999.]

(18 ,36 ,54 ,72 ,90 ,108 ,126 ,144 ,162 ,180 ,198 ,216 ,234 ,252 ,270 ,288 ,306 ,324 ,342 ,360 ,378 ,396 ,414 ,432 ,450 ,468 ,486 ,504 ,522 ,540 ,558 ,576 ,594 ,612 ,630 ,648 ,666 ,684 ,702 ,720 ,738 ,756 ,774 ,792 ,810 ,828 ,846 ,864 ,882 ,900 ,918 ,936 ,954 ,972 ,990 ,) >>Total =  55 numbers that are multiples of 6 AND 9.

-3.5x + 28 = 231     subtract 28 from both sides

-3.5 x = 203            divide both sides by  - 3.5

x = -58

Thank you! Could you please explain farther?

300 students   with   19 out of 50 qualifying means    114 qualify

only 75 are picked via the lottery

which means 39 qualifiers do NOT get picked  

39/114 = 13/38 =  0.3421052631578947                         (I think)

Maybe this:  (as I am not sure what the question is asking)

probability that you are in is  114/300  

probability that you LOSE the lottery   39 / 114

multiplied together =   39/300 = 13/100

Here is a Venn Diagram of this situation;

sum(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100) = 2,550

The x-intercept is (-18*3/7,0) = (-54/7,0).

Florida issues license plates in which the first three and last slots are filled with letters, and the fourth and fifth are filled with digits. Thus, there are \(26^4 \cdot 10^2 \) Florida license plates possible. North Dakota, however, issues license plates in which the first three slots are filled with letters and the last three slots are filled with digits. There are thus \(26^3 \cdot 10^3\) possible North Dakota license plates. Multiplying these out and taking the difference yields an answer of \(\boxed{28121600}\).