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https://web2.0calc.com/questions/right-triangles_35

We can write an equation. 

The measure of an interior angle is \( (n - 2)(180^\circ) / n \) where n is the sidelengths. Since, the exterior angle is 180  minues the interior angle, we have

\( (n - 2)(180) / n = 2 · ( 180 - (n - 2)(180) / n ) \\ (n - 2)(180) / n = 360 - 2 · (n - 2)(180) / n ) \\ 3 · (n - 2)(180) / n ) = 360 \\(n-2)(180)/n=120 \\(n-2)/n=2/3 \\n-2=2n/3 \\3n-6=2n \\n=6\)

So our answer is 6. 

We can check our work.

A 6-sided polygon has an interior of 120 and exterior of 60.

Thank! :)

A  = 6(1/2) s^2  sqrt (3) / 2

3/2 = (3/2)   sqrt (3)  s^2

1 = sqrt (3)  s^2

1/sqrt 3  = s^2

s =  sqrt (1 /sqrt 3)  =  1 / ⁴√3 

One way to deal with the restriction is to solve the problem as if it didn't exist and then subtract the solutions we can't have.

There are 6 children and 6 seats. The 1st child has 6 choices. For each of those 6 options, the 2nd child has 5 choices, because the 1st child occupies a chair. For each of those 5 options, the 3rd child has 4 choices, and so on. So, there are 6*5*4*3*2*1 options. One way to see this is to draw a counting tree. 6*5*4*3*2*1 is more commonly known has 6 factorial, or 6!. 6! = 720, which is the number of ways you can sit the children without restrictions. However, no two siblings can sit in front of their sibling. This removes 3 options, so the answer is 720 - 3 = \(\boxed{717}\).

The youngest kids both get 4 pieces, so we subtract eight pieces from the total and continue, ignoring this. There are 5 kids and 5 pieces. Each piece of candy can go to each kid and has 5 choices. If each piece has 5 options, then there are 5*5*5*5*5, or 5^5, ways to distribute the candy. 5^5 = \(\boxed{3125}\), which is the answer.

B) We can start the same way that we solved A). However, now we are dealing with the columns, and we have an extra counter. So, each column needs 3 counters. Instead of counting the number of ways we can place the counters, let's instead count the ways we can pick the empty square. We can pick each square in the column to be empty, so there are 4 ways. 4 ways to do this in each column, which is 4*4=16. So, the answer is \(\boxed{16}\).

A) The square has four columns and four rows. Each row and column has 4 squares. In each row, we have four squares to place the counters in each row, or 4 choose 2, which is \(\frac{(4\cdot3\cdot2\cdot1)}{(2\cdot1)}\). This equals 12. So, there are 12 ways to place 2 counters in each row. However, there are 4 rows, so we multiply by 4. 12 times 4 is \(\boxed{48}\).

Thanks! :)

I like your polynomial  approach  !!!!!

At least 5 heads means the same as no more than 4 tails

0 tails = C(12,0) =  1 possible sequence

1 tail  = C(12,1) =    12 possible sequences

2 tails =  C(12,,2)  =   66 possible sequences

3 tails  = C(12, 3)  = 220 possible sequences

4 tails  = C (12,4)  =  495 possible sequences

At least 5 heads =   sum of 12rh row of Pascal's Triangle - ( 1 + 12 + 66 + 220 + 495)  =

2^12 - ( 1 + 12 + 66 + 220 + 495)  = 

3302 possible sequences

There are actually more 2-digit numbers that are cozy than not. 

We can just list them out!

\(11,13,15,17,19 \\ 31,33,35,27,39 \\ 51,53,55,57,59 \\ 71,73,75,77,79 \\ 91,93,95,97,99\)

These numbers are those who are not cozy. Counting through them, there are 25. 

In total, there are 90 two-digit numbers, so we have

90 - 25 = 65. 

So 65 two-digit numbers are cozy. 

Thanks! :)

Factor 2000  =  2^4 * 5^3 =  1* 4^2 * 5^3 =   1 * 4 * 4 * 5 * 5 * 5

Possible identifiable sequences  =    7!  / [ 2! * 3! ]  =    420

Although I don't really recommend it, we could just list out all the numbers since groups 5 and 4 are really not that big.

We have 

Group 4 - 26, 28, 30, 32, 34, 36

Group 5 - 38, 40, 42, 44, 46, 48, 50

Adding the numbers in group up, we get 308, which is our answer. 

We could also generate a polynomial equation to help us. 

We could put it in the form of a polynomial with 3rd degree in ax^3+bx^2+cx+d

From the problems above, we can get the system

\( a + b + c + d = 12\\ 8a + 4b + 2c + d = 44\\ 27a + 9b + 3c + d = 100\\ 64a + 16b + 4c + d = 186\\\)

Solving this, we get

\(a =1 \\ b= 6 \\ c= 7 \\ d= -2\)

This means we have a polynomial \(n^3 + 6n^2 + 7n - 2\). 

Plugging in 5, we get 308.

Thanks! :)

Well, notice that LCM[6, 3], is 6, so every factor of 6 between 0 and 100 is a factor. 

We have \(100/6=16.67\), rounding down means there are 16 factors. 

However, this is not including 0, so we add 1 to get 17. 

So both Sarah and Amos say 17 numbers out loud. 

Thanks! :)

This triangle is impossible to exist. 

\(144^\circ + 66^\circ = 210^\circ\), which is more than 180 degrees, what the triangle SHOULD add up to. 

Thanks! :)

ab -3a + 4b  = 131        take the product of the coefficients on a,b.....add this product to  both sides

ab - 3a + 4b + (4 * - 3)  = 131 + (4 * -3)

ab - 3a + 4b - 12  =  119            factor the left side

(a + 4) ( b - 3)  = 119

Factors of  119  are    1   7    17     119

Using   17 and  7

( 13 + 4) ( 10 - 3)   →  l a - b l  =  l13 - 10 l =  3

https://web2.0calc.com/questions/quadratic_84013

We can use the Law of Cosines

BC^2 = AB^2 + AC^2  - 2 ( AB * AC) * cos (120)

(x + 4)^2  = x^2  + (x + 2)^2  - 2( x * ( x + 2)) (- 1/2)    simplify

x^2 + 8x + 16 =  x^2 + x^2 + 4x + 4  + x^2 + 2x

x^2 + 8x + 16  = 3x^2 + 6x  + 4

2x^2 - 2x - 12 =  0

x^2 -x - 6  =  0

(x -3) ( x + 2)   =  0

X -3   = 0

x = 3

https://web2.0calc.com/questions/angle-problem_3#r1

3 hrs 36 min =   3 + 3/5  hrs  = 18/5 hrs

3hrs 45 min = 3 + 3/4 hrs =  15/4 hrs

Let the distance between the towns = D

So....the round trip  = 2D

Let the speed of the plane in still air = S

With the wind the the plane travels at S + 30  mph

Against the wind the plane travels at S - 30 mph

So...in still air we have this equation using Distance / Rate = Time

2D / S =  18/5

2D = (18/5)S

S = (10/18)D   =  (5/9)D   (1)

With the wind blowing we have this 

D / (S + 30)   +  D/ (S -30)  =  15/4      (2)

Sub (1)  into (2)

D / (5/9 D + 30)  + D /( 5/9 D - 30) = 15/4     simplify

D(5/9D -30)  + D(5/9D + 30)  = (15/4) (5/9D + 30) (5/9D -30)

(10/9)D^2  = (15/4) ( 25/81 D - 900)

(10/9)D^2  =  (125/108)D^2 -3375

(125/108 - 10/9)D^2  = 3375

(125 / 108  - 120 /108)D^2  = 3375

(5/108)D^2  = 3375

D^2  =  108 / 5 *  3375

D^2  = 72900          take the square root

D = 270  miles

Let AE = "a"

Note that angle AGE = 90   because GE is perpendicular to AF

Let angle  GAE =  theta

Because AF is parallel to EC , then angle GAE  =  CED

sin ( GAE)  =  1/a

tan (CED)  = 8 / (10 - a)

sin (CED)  =   8 / sqrt [ 8^2 + (10-a)^2]  =  8 /sqrt [ a^2 -20a + 164]

sin (GAE)  = sin (CED)

1/a = 8 / sqrt [a^2 -20a + 164 ]       square both sides and  cross-multiply

a^2 -20a + 164  =  64a^2

63a^2 + 20a - 164 = 0

a ≈ 1.4625

The strip is a parallelogram with a height of 8 and a width of a = 1.4625

Area =  8 * 1.4625 =  11.7 

Note something interesting.......the numerical value of the area ≈ the numerical value of the length of  AF, EC

-1 < 2x < 12    divide through by 2                       -1 < 3x - 7 < 12      add 7 to each

-1/2 < x < 6                                                           6  <   3x < 19      divide through by 3

                                                                              2 <  x   < 19 / 3

Most restrictive interval

2  <   x  <  6

https://web2.0calc.com/questions/bisectors_25

https://web2.0calc.com/questions/right-triangle_44

Break this into two parts

1 - 3x < x + 8                   x + 8 ≤  7 - x

-7 < 4x                             2x ≤  -1

x > -7/4                              x ≤ -1/2

( -7/4 , -1/2 ]