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The ratio of four sides of trapezoid is 

2:1:1:1

The area of the trapezoid: 27^sqrt(3) 

Well           Sum

2   :1   :1   :1   :5

\(x_1\)  \(x_2\) \(x_3\):\(x_4\)=\(27*\sqrt{3}\)

\(x_1=18.7 \)

\(\)\(x_2=x_3=x_4=9.35\)

Median:

9.35,9.35,9.35,18.7 

The median is (9.35+9.35)/2=9.35 units 

سته و اربعه على سالب ثلاثة يساوي كم

Six and four by minus three is equal to how many?

Sechs und vier mal minus drei sind gleich wie viele?

مرحبا ضيف

\(6+4\cdot(-3)=\color{blue}-6\)

   !

6+4-3=7

As follows:

If tan⁡〖θ=8/15〗 and π≤θ≤3π/2, find the values of the other 5 trig functions.

Hello subarashiib!

\(\color{BrickRed}tan (\theta)=\frac{8}{15}\\ cot(\theta) =\color{blue}\frac{15}{8}\\ sin(\theta)=\pm \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}}=\pm \frac{\frac{8}{15}}{\sqrt{1+(\frac{8}{15})^2}}=\color{blue}\pm0.470588 \)

\(cos(\theta)=\pm\frac{1}{\sqrt{1+tan^2(\theta)}}=\pm \frac{1}{\sqrt{1+(\frac{8}{15})^2}}=\color{blue} \pm0.8823529\\ sec(\theta)=\pm\sqrt{1+tan^2(\theta)}=)=\pm\sqrt{1+(\frac{8}{15})^2}=\color{blue}\pm1.1\overline{33}\\ csc(\theta)=\pm\frac{\sqrt{1+tan^2(\theta)}}{tan(\theta)}=\pm\frac{\sqrt{1+(\frac{8}{15})^2}}{\frac{8}{15}}=\color{blue}\pm2.125\)

  !   \(\small asinus\)

Given that f(x) = x^k where k < 0, what is the range of f(x) on the interval [1, ∞)? Please explain, I want to understand how the answer is gotten. 

Ok, I will try to explain. 

\( f(x) = x^k \;\;\;where\;\;\; k < 0, \;\;\:and \;\;\;\ x\ge1\\ let\;\; n=-k \;\;\;so\;\;\;n>0\\ f(x) = x^{-n}=\frac{1}{x^n}\\ now\;\; x\ge1 \;\; and\;\; n>0\;\;\;\\so\\ x^n\ge1\\ so\\ 0< \frac{1}{x^n} \le1\\ 0< x^{-n}\le1\\ 0< x^{k}\le1\\ \)

Here is the graph

I have not done a full answer, and you will need to checke what I have done very carefully.

Here is my outline.

There are lots of angles you can work out on this diagram.  You were given 30 degrees as one of the rhombus angles, so the others must be 30, 150,150.

Since the sides of all the given squares is 6. 

    The distance from a vertex to the centre of its square is 3 sqrt2  (pythagoras's theorem)

All the green triangles are congruent.

Therefore the new quadrilateral is a square.

Using cosine rule I got side squared = 54 units squared. 

thank you so much!

Don't worry if you don't get back to me in time, I found that the recommended taper-ratio for commericial airplanes is 0.4.

However, a more specifically recommended taper-ratio for balsa gliders would help :)

Jenny, what you're describing is called a Tangram Vanishing Piece Puzzle. 

You can find a discussion (solution) of them here:  https://www.archimedes-lab.org/workshoptangram3.html     

_.

jenny, please read what i wrote:  you could not allow the apparatus to turn sideways.

Unfortunately, all of these answers seem to be wrong.

Thanks

Thanks

1. Angle D is 135 degrees.

2. Arc AOB is 240 degrees.

Second one

center  =  ( 1,6)

focus  = (-1, 6)    means that c   = 2

vertex  (6,6)   means that a  =  5 ⇒   a^2 = 25

And b^2  = a^2 - c^2 =  5^2 - 2^2  = 21

So....the equation is

( x - 1)^2            ( y - 6)^2

_______   +     _________   =     1

    25                     21

First one.....the major axis lies along y

So

24  =  a^2

sqrt (24)  = sqrt (4 * 6)    =  s sqrt (6)  =  a

The major axis   = 2a =  2  (2sqrt (6) ) =   4sqrt (6)

The minor axis lies along  x

So

12 =  b^2

sqrt (12)  =sqrt ( 4 * 3)  =  2sqrt (3)  = b

The minor axis  = 2b = 2 ( 2 sqrt (3)) = 4 sqrt ( 3)

Total cost  =  Number of yards bought x  cost per yard  =  4.25  x 1.79......can you  take it from here   ????

Second one

The vertex of a parabola is (6,2) , and the equation of its directrix is y = 4.

What is the equation of this parabola in standard form?

The directrix is above the vertex.....therefore.....this parabola  turns downward

a = the distance between  the vertex and the  directrix  = 2

And the vertex  = (h, k) = ( 6, 2)

So....the form is 

(x - h)^2  = -4a ( y - k) 

(x - 6)^2  =  -8 ( y - 2)  

First one  :

x^2 - 6x  -  2y  + 7   = 0        rearrange as

2y - 7  =   x^2  -  6x                  complete the square on  x

Take (1/2)  of 6  = 3....square it  = 9.....add to both sides

2y - 7+ 9    =  x^2  - 6x + 9            factor the  right side, simplify the left

2y + 2   =  ( x - 3)^2           r

(x - 3)^2  =  2y + 2

( x - 3)^2  = 2 ( y + 1)

x + y = 9/2, 2x^2 y^2 - 13xy + 18 = 0

2x^2y^2  -  13xy  +  18   =    0    factor as

(2xy  -  9 ) (xy - 2)   =  0

Settting each factor to 0  and solving we have

2xy - 9  = 0                xy  - 2  = 0

xy  = 9/2                    xy  =  2

y = [9]/ [ 2x]                y  = 2/x

And using  x + y  = 9/2   we can solve this

x + [9] / [2x]  = 9/2       multiply through by 2x

2x^2  + 9 =  (9/2)(2x)    rearrange

2x^2  - 9x + 9   = 0

(2x - 3) ( x - 3)  = 0

So

2x - 3  =0        x - 3  = 0

x = 3/2              x   =  3

And y  = 9 ] [ 2(3/2)]  = 3      And    y  = 9/ {2*3] = 3/2

So....two solutions are    ( 3/2, 3)   and  ( 3 , 3/2)

And for the other solutions, we have that

x + 2/x  = 9/2       multiply through by x

x^2 + 2 = (9/2)x

2x^2 - 9x  + 4  = 0

(2x - 1) ( x - 4)  = 0

2x - 1  = 0              x  - 4  =0

x =1/2                   x  = 4

And y  = 2/ [1/2]   =  4        And  y = 2 / 4   = 1/2

So....the other two solutions are   (1/2, 4)   and ( 4, 1/2)

 ​ 

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6/9   =  [6 * 8] / { 9 * 8]  =   48/72

7/8  = [ 7*9] / [ [ 8 *9 ] =   63/72

So...."C" looks to be correct

If A = 3^x + 3^{-x} and B = 3^x - 3^{-x}, then find A^2 - B^2.

A^2  =  ( 3^x + 3^(-x) )  ( 3^x + 3^(-x) )   =  [ 3^(2x) + 2 (3^x * 3^(-x) ) + 3^(-2x)  ]  =

[ 3^(2x)  + 2  + 3^(-2x) ]

B^2  =  ( 3^x - 3^(-x) )  ( 3^x - 3^(-x) )   =  [ 3^(2x) - 2 (3^x * 3^(-x) ) + 3^(-2x)  ]  =

[ 3^(2x)  - 2  + 3^(-2x) ]

So

A^2   - B^2  = 

[ 3^(2x)  + 2  + 3^(-2x) ] -  [ 3^(2x)  - 2  + 3^(-2x) ]  =

4