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ok thanks

thanks! correct!

Yep.

But make sure that A < B < 23 < C < D too. :))

=^._.^=

x + 15 + x - 15 + x + 30 = 180

3x + 30 = 180

3x = 150

x = 50

=^._.^=

90 + 3x + 4x - 1 = 180

89 + 7x = 180

7x = 91

x = 13

=^._.^=

lets see

we have

A  B   23  C D

the mean is 20 so that means they add to   5*20 =100

100-23 = 77

So choose any 4 numbers that add up to 77

There are 2 cases. 

Penelope and Quentin are not officers. 

20*19*18 ways to choose from the other 20 people. 

Penelope and Quentin are officers. 

3*2*20 ways to choose another person from the other 20 and give assign them spots. 

20*19*18+3*2*20

=^._.^=

Draw this out and create a right triangle. 

With the right triangle, you should be able to figure out the other 2 angle measurements. 

Then, use the law of sines to find the height from the ground. :))

=^._.^=

Next time, just one question per post, thanks  

Your answer is right but your logic is wrong.

10 ways to pick the girl, 10 ways to pick the boy But the girl can be president or vise, once the girls' position is fixed so is the boys so that is 2 ways

answer  10*10*2 = 200

6 vowels

6/26  

It will be x minutes past 9

Big hand  360/60 * x = 6x  degrees  (past 12)

Little hand = -90+x/2 

difference =  6x- (-90+x/2) = 6x+90 - x/2 = 90+ 11x/2

90+11x/2 = 139

11x/2       =  49

11x          =  98

x =8.909090....

x approx = 8 min  55 sec

It is approx  8 minutes and 55 sec past 9

I just checked.

None of your questions have been whited out.

Maybe it was a temporary glitch.

If you click on your username and then on 'my profile' you will get access to all of your posts.

(Maybe not if they have been whited out or hidden but none of yours are whited out or hidden)

Nice answer Ginger 

No guest, ur wrong for 2 its 1440 

first ones 20 , 20 people, 10 people of each gender, 20*10=200 ways... Figured it out

Hi i replied on ur initial post.

Hello, 

This problem is really challenging. Could you tell me where you got it from if u know?

Anyway, i got the n*s=50, is it correct? If so, tell me and I will explain how I got it.

TP is a line that tangent to a circle centered at O. If PQ∥TO and ∠OTP=28∘, find the measure of ∠POQ in degrees.

This question has illusion of complexity because it’s intuitive for many mathematicians to find the values of the two (2) numbers (a) and (b), but the question only asks for their sum. 

Solution:

\(\small \text{ }\\ \begin{align*} \small x^2 + y^2 &= 327\\ \hspace{1em} \small x + y &= 101\\ \end{align*} \;\\\;\\ \text {Note that }\\ \left(x+y\right)^2 \;\; = \quad x^2 + y^2 +2xy \hspace{3ex}| \scriptsize \text { perfect square expansion formula }\\ \text{then}\\ \left(x + y\right)^2 \;\; = \quad 327 +2(101)\\ \left(x + y\right)^2 \;\; = \quad 529\\ \text{then }\\ \sqrt{\left(x + y\right)^2} \;\; = \quad \sqrt{529}\\ \hspace{1.5em}{\left(x + y\right)} \;\; = \quad23 \; \hspace{2em} \leftarrow \small \text{sum of (x) and (y)}\)

GA

Perimeter of rhombus = 4 × side = 148                        A                        B

⇒ side = 37 

Consider rhombus ABCD                                                     O

      AB = BC = CD = DA  &  AC ⊥ BD

 AC & BD intersect at O.                                       D                       C

BD = 20  so, OB = OD = 10

Since ∠COD = 90             [diagonals of rhombus intersect at right angles]

⇒ OC = 35.6

Diagonal AC =  71.2 units 

ABC is the right triangle, AC and BC are the sides and AB is the hypotenuse. 

AC = 12 and BC = 5    ⇒ AB = 13 

⇒ Area of △ABC = 1/2 × AC × BC 

                            = 1/2 × 12 × 5

                            = 30

also, area of △ABC = 1/2 × CH × AB

                                = 1/2 × CH × 13

1/2 × CH × 13 = 30 

⇒ CH = 60/13  units  = 4.6 units 

The following graph should help:

The balls are labeled 2, 4 and 8. A ball is picked, the value on the label is recorded and then the ball is returned to the bag. This will be done three times and then the values will be mutiplied. 

So tha sample space = {(2,2,2), (2,2,4), (2,4,4), (4,4,4), (2,2,8), (2,8,8), (8,8,8), (2,4,8), (4,8,8), (8,8,4),so on....}         (27 permutations)

⇒ Sum of distinct products = (2 × 2 × 2) + (2 × 2 × 4) + (2 × 4 × 4) + (4 × 4 × 4) + (4 × 4 × 8) + (4 × 8 × 8) + (8 × 8 × 8) + (8 × 2 × 2) + (8 × 8 × 2) +(2 × 4 × 8)

                                            = 8 + 16 + 32 + 64 + 128 + 256 + 512 + 32 + 128 + 64

                                            =  1240

Sum of all distinct products is 1240.