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I don't know how to solve it mathematically, but a short computer code gives the following number:

190 / 6^5 =2.4434 %. 

Let us if a mathematician can arrive at the same number using permutations and combinations.

Since the point   (0, -1) is on the graph, c   must  be  -1

Let the long diagonal  of the rhombus    =  L

By the intersecting chord theorem  we have that

(3/2)r * (1/2)r    = [ ( 1/2) L ]^2

(3/4)r^2  = (1/4)L^2       multiply both sides by 4

(3)r^2  =  L^2       take the square root

(√3) r  = L

So.....the area of the rhombus  =   (1/2) product of the diagonals   =

(1/2) (r) (√3) r  =

[√3/2] r^2

2) In terms of pi, what is the area of the circle defined by the equation 2x^2+2y^2+10x-6y-18=0

Divide every term by 2  and rearrange as

x^2 + 5x  + y^2  - 3y  =  9      complete the square on  x  and y

x^2 + 5x + 25/4  +  y^2 - 3y + 9/4   =   9 + 25/4  +  9/4        factor and simplify

(x + 5/2)^2  +  ( y - 3/2)^2   =  36/4 + 25/4 + 9/4

(x + 5/2)^2  + ( y-3/2)^2  =  70 /4  = 35 / 2   =17.5

The area  =  pi * r^2   =    pi * 17.5    =  17.5 pi

1)

Since (0,0)  is on the graph, c   = 0

So we have that

a(3)^2  + b(3)  = -3     ⇒  9a  +  3b  = -3   ⇒    -3a - b  =  1    (1)

a(1)^2  + b(1)  =  3     ⇒   a  +  b    =  3           (2)

Add(1)  and (2)  and we get that

-2a  =  4

a = -2

b = 5

100 a +  10 b +  c   =  -200 + 50  + 0   =   -150

Great question!  a = 1, b = 2, and c = 4, so a - b + c = 3.

a =  40°   = c

b =  180 - 2(40)  = 100°

e  is supplemental to  c  =  180  - 40  =  140°

So

a + b + c + e  =

40 + 100 + 40  + 140  =

320° 

Large square side =>   a = 12cm             A = [ b*sqrt(2) + a*sqrt(2) /2] * d/2

---- II ---- diagonal  =>   d

Small square side  =>   b = 6cm                                                 

Green area             =>  A                                               A = 144 cm²  

Do your parents allow you to go into the STEM path?

2)

\( f^{-1}\left(f^{-1}(50)\times f^{-1}(10)+f^{-1}(26)\right) \)

f^-1 (50)  =  7

f^-1 (10)  = 3

f^-1  (26)   =  5

So   we are evaluating this

f^-1  (  7  x 3  +  5)  =

f^-1  ( 21  + 5)  =

f^-1 (26)  =

5

1)  

We have these equations

(-1 -h)^2  + (2 - k)^2  = r^2

(3 - h)^2  + k^2  = r^2

(9 - h)^2  + k^2  = r^2

Subtract the  second equation from  the 3rd and we have that

(9 - h)^2  - (3 - h)^2  =  0    simplify

h^2 - 18h  + 81  -  [ h^2 - 6h  + 9  ]  = 0

-12h  + 72   =0

-12h  = -72

h = 6

And  equating the first two equations

(-1 - 6)^2  + (2 - k)^2  =  (3 - 6)^2  + k^2

49  + k^2 - 4k + 4  =  9  + k^2

53  - 4k  =  9

44 - 4k  = 0

k  = 11

So

(h, k) =  (6, 11)

And  h + k  =  17

Music will be challenging and sometimes annoying, I know cus I play the oboe and sometimes I don't even like playing it. I would just play an instrument as a hobby and do computer science.

Circle's radius     =>  r                   A = [r/tan(q/2)*r]*3   

Triangle's angle  =>  q

Triangle's area    =>  A                                  A = 11.69 u²  

\( $f(x)=\dfrac{2}{x+1}$ \)

\( $f^{-1}\left(\frac{1}{5}\right)$ \)

Note that  we  need  to  solve this

(1/5)  =   2 / ( x + 1)          and we can write

5   =  ( x + 1) / 2             multily both sides by 2

10  = x +  1                   subtract 1 from both sides

9  =  x  =  f^-1(1/5)

So  the point (9, 1/5)  is on f(x)     and the point  (1/5 , 9)  is on f^-1 (x)  

Rip but my parents are forcing me to do orchestra nuuuuuuuuuuuuuuuuuuuuuuu

Why do something that you hate  [ if you're not required to ]   ????

Follow your heart......it will rarely fail you.....

\( $\frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}}$ \)

Express in form

\($\frac{a^2\sqrt[5]{b}+b\sqrt{a}}{ab}$\)

Note that  (16)^(1/5)   =  (2^4)^(1/5)  = 2^(4/5)

So...multiply the numerator and denominator of the first fraction by 2^(1/5)

Multiply  the  numerator /denominator of the second fraction by  3^(1/2)

So  we have

3 * 2^(1/5)           3^(1/2)

__________  +    _______          get  a common denominator

       2                     3

3 * 3 * 2^(1/5)   + 2 * 3^(1/2)

_________________________       simplify

               2  *  3

3^2 *  ⁵√ (2)      +   2 * √(3)

___________________________

                3 * 2

a =  3      b  = 2 

a +  b   =  5

Strictly by  "educated guessing"

Let us assume that  right triangle ABC  is a Pythagorean multiple  of a 3 - 4 - 5    right triangle  such that

BC = 36

AB = 48

AC  = 60

And let us assume that  right triangle  DBC  is  a Pythagorean multiple  of  a  5 -12- 13  right triangle  such that

DB = 15

BC = 36

DC = 39

Then BE  =  √ [AE^2 - AB^2]  =  √  [ 52^2  - 48^2]  = √400  =   20

So

DE = √[ DB^2  +  BE^2  ]  =  √ [ 15^2  + 20^2 ]  = √ 625   =  25

Here is the image to show  that this is true  :

2a + 3b  = c   ⇒        2a + 3b  - c  =  0          (1)

a + b + c   =  42  ⇒   -3a -3b - 3c =  -126        (2)

c - a  = 19     (3)

Add (1) and (2)

-a - 4c  = -126       ⇒   a + 4c =  126    (4)

Add (3)  and (4)

5c  =  145

c = 29

c - a = 19

29 - a  = 19

a = 10

a + b + c  =42

10 + b + 29  = 42

b = 3

So

(c + 1) / [ ab  ]  =     (29 + 1) / [ 10 * 3 ] =     30 / 30    =   1

In the given figure, ABCD is a parallelogram. If the area of the parallelogram is 156 sq cm, find the length of AL.

Hello Guest!

\(13cm\cdot \overline{DL}=156\ cm^2\\ \overline{DL}=156\ cm^2/13\ cm\\ \overline{DL}=12\ cm\)

\( \overline{AL}=\sqrt{(13\ cm)^2-(12\ cm)^2}\)

\( \overline{AL}=5\ cm\)

  !

Find the number of positive real numbers x such that 

\(\log_{10} x = \sin x \)

Hello Guest!

\(\log_{10} x = \sin x\\ 10^{sin\ x}=x\\ 10^{sin(x)}-x=0 \)

x=2.696256562..  (determined graphically)

  !

Suds,

You asked a question.

You have had a stack of people spending a lot of time helping you.

You give them no credit and no interaction.

Then you delete your question.

Do you realize how incredibly rude that is?

u r w

sec (4A)  =  csc (A - 20)

These  are cofunctions which means that    

(4A ) +  (A - 20) = 90

5A  =  110

A = 22°

x^2 + y^2  + z^2  =  18          xy   +  yz  +  zx  =  9

x^2  +  y^2  =  18   -  z^2

Here are the possibilities   for  x, y  and  z   (in any order)

x         y         z

1         4        1

-1       - 4      -1     

3         3        0

-3        -3       0

So

l  x  l    +  l y l    +  l z l  =   6