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Thanks, tertre....!!!

24*2/3=24/3*2=16

Note......

When θ = 0     r = 3

When θ = pi / 6, r = 0

And this traces out   (1/2) of a "leaf"

So....each pi/6 rads will do the same

So.....6 of these will trace out 3 full "leaves"

So

6 (pi/6)  =   pi  =    M   =   (180”)

Just count the end results. Since we have two for each, our answer is \(2*2=\boxed{4}.\)

Type 5 x 10^300   =   = 

Just multiply: \(\frac{5}{7}*14=5*2=\boxed{10}.\)

r = √ [ x^2 + y^2 ]

cos theta =   x/r =  x / √ [ x^2 + y^2 ]

sin theta = y/r = y / √ [ x^2 + y^2 ]

So we have

√ [ x^2 + y^2 ]   =                 7

                                  _____________

                                    [  3x +  2y ]

                                   ___________

                                   √ [ x^2 + y^2 ]

√ [ x^2 + y^2 ]  =     7 √ [ x^2 + y^2 ]

                              _____________

                                    3x   + 2y

1     =          7

              _______

               3x + 2y

3x + 2y  =   7

In the form Ax + By = C.....the slope is     -A / B

So.....slope   =        -3/2

P(6,2)  = 30

Good Job, Tanmai......!!!!!

"Of" means "multiply"

(2/3) * (15/1)  =  30 / 3  =  10

Let A = drawing a spade from 52 cards

P ( drawing a spade ) = 13/ 52 =  1 / 4    = P(A)

Let B = probability of drawing a spade (after replacement) = 13/52 = 1/4 = P(B)

So

P ( A and B ) = P(A) * P(B)  =  (1/4) * (1/4)  =  1 / 16

6 shows.....choose any 4  =

C(6,4)  =   15   different combos

B. 4

Why is math what? Fun? Math is fun because of the satisfaction from getting the right answer.

Here is me trying:

There are 6 letters, right? All of them are different. You make a tree diagram or a chart.

There are this many: FA, FB, FR, FI, FC, AF, AB, AR, AI, AC, BF, BA, BR, BI, BC, RF, RA, RB, RI, RC, IF, IA, IB, IR, IC, CF, CA, CB, CR, CI.

That is if the same letter cannot be used twice within one permutation. If that can happen, then add 6 to the number.

My answer: 30

See the answer here:  https://web2.0calc.com/questions/help_53845

The cut must be a diagonal cut if two triangles are formed

The 13 inch cut  forms the hypotenuse of aright triangle....and the length of the paper forms one of the legs

The width is given by

√ [ 13^2 - 9^2 ]    =  √ [ 169 - 81 ]  =   √ 88   ≈  9.38 inches

Ah, yes.....disregard my answer....!!!

Given: Parallelogram MNPQ, M(3a-b,b) N(3a+b,3b) P(6a+4b,3a+3b)

Find: The coordinates of point Q

The opposite sides are parallel

So....the slope between  MN is 

[ ( 3b - b) ]                           2b

_________________  =   _____  

[ (3a + b) - ( 3a - b)]            2b

And the slope joining PQ is the same....let (x,y) be the coordinates of Q

[ (3a + 3b)  - y]                      2b

__________________ =   ______ 

[ (6a + 4b) - x ]                      2b

This implies that  x = (6a + 2b)  and y = (3a + b)

So Q =   [ 6a + 2b , 3a + b ]

We can check this..since NP and MQ are parallel.....the slope of NP should = the slope of MQ

NP 

[ (3a + 3b) - 3b]                    3a                    a

_________________ =     _______  =      ____

[(6a + 4b) - (3a + b)]           3(a + b)            a + b

MQ

[ b - ( 3a + b) ]                          - 3a         -3a                      a

_________________  =        _____ =   ________   =    _____

[ (3a - b) - (6a +2b) ]             -3a -3b      -3 ( a + b)          a + b 

Hint: Try and find criteria that helps narrow down answers.

From the graph you can see that it is a (4 x 3) rectangle with two (right) triangles attached  (base 4   h  2)    

You have been shown how to do several of these similar graphs.....try to figure out an answer and get back to me if you have questions......

The triangles will be RIGHT triangles with leg lengths of 8 and 12

Pythagorean theorem will give the length of the hypotenuse (which is the diagonal cut)

sqrt (8^2 + 12^2) = 14.42 in

The perimeter of ONE of these triangles is the sum of the sides    12 + 8 + 14.42 = 34.42 inches

most of the time you end up with the square root of a negative number.

for n>4 you always do.

CPHILL, the problem states that there is a mode. The angles should also be integers.