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Note  that  triangle  UVW  is similar to triangle UWT

Therefore

UV  / UW  =  UW / UT

4/ UW  = UW   /  6.25

So

UT * UV  =  UW * UW 

6.25 * 4  =  UW^2

25 =  UW^2     take  the positive root

5  = UW

2x+1=3x

1 = 3x - 2x

1 = x

x = 1

It's wrong though...

Thank you!

 ∛32  =   ∛4  *  ∛8  =   ∛4  *  2

So  we  have

      2

__________     =

 ∛4  +  ∛32

       2

__________    =  

 ∛4  + 2 ∛4

          2

    ___________   =

     ∛4  ( 1 + 2)

          2

 ____________                  multiply  top / bottom  by      ∛2

       3  ∛4   

          2  ∛2

_____________    =

       3  ∛4 *  ∛2

         2 ∛2

     ________   =

          3 *  ∛8

          2  ∛2

       ________   =

          3  *  2

               ∛2

            ____             A  +  B  =    5       CORRECTED  !!!!

                3

Thank you guys!

Use Heron's formula     p = (13 + 14 + 15)/2 = 21

area = sqrt ( p (p-13)(p-14)(p-15)  )  

         = sqrt (21 (8)(7)(6)  )  = 84    units²         as Penguin stated !

Thanks . However I do not understand if if there is a specific equation that should be used?

I think it is 5/6, but I am not very confident with the answer I have given. 

I'm pretty sure its 84

2x+1=3x

-2x     -2x

1= 1x

x=1

The  horse  can  cover one  area of  3/4  of  a circle  with a radius of  15m   = (3/4)pi (15)^2

Additionally  the horse  can  cover  the area of  two quarter-circles with  a radius  of 3m  =  pi (3)^2 / 2

So....the   horse   can    graze in an area of   pi  [ 15^2 * 3/4   + 3^2 / 2  ]  ≈ 544.28 m^2

N       D        Q

9

7       1

5       2

3       3

1       4

2       1        1

4                 1

         2        1

120 + 130x = 180 * (x+1-2)

120 + 130x = 180x - 180 

300 = 50x

x = 6

You get that x = 6, but there is still one angle that equals 120, so you have a 7 sided polygon.

Borrowing on the  guest's answer....the GP  is

3   15    45    135    405

The  number of grey tiles   can  be found thusly  :

Let  n  be  the  number  of   rows of tiles

The  number of  yellow  tiles   =  n + (n -1)    =  81

So

2n  -  1   =  81

2n =  82

n = 41    rows

Number  of    grey tiles   = ( n - 2) ( n - 1)  + (n - 1)  =   ( n - 1) ( n - 2 + 1)  = ( n - 1) ( n -1)  =

(n -1)^2   =   (41   -1) ^2  =   40^2   =   1600

11^(8^5) = 11^(2^15)

2^15 = 32768, so we are finding the number of digits of 11^(32768) in base 11. The answer is simply 32768.

\(\frac{1000000}{\frac{100000+150000}{2}} = \frac{1000000 \cdot 2}{250000} = \boxed{8}\)

As a hint, the sum of angles in a triangle is 180 degrees.

\(\sqrt{\pi x^4} = \boxed{x^2 \sqrt{\pi}}\)

\(20+40 = \boxed{60}\)

Let \(z = a + bi \)

\((a^2-b^2) + 2abi + 2a + 2bi = 31 - 8i\)

Separate real and imaginary components.

\(a^2-b^2 + 2a = 31\)

\(2ab+2b=-8\)

The base is 24 centimeters, so the legs are 28 centimeters each. So, the perimeter is 24 + 2(28) = 80 centimeters.

The height of the triangle is \(\sqrt{28^2 - 12^2} = 8\sqrt{10}.\) That makes the area \(\frac{1}{2}(8\sqrt{10})(24) = \boxed{96\sqrt{10}}\) cm^2.

45 *r     * r   = 405

r = 3

Let n be the number of students. We know that

n = 2 (mod 4)

n = 1 (mod 5)

15 students are girls, so there is a maximum of 29 students.

One solution for n is 26. So, the teacher has 26 students.