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I see. Thanks!

Thanks 

We have  C(6,2)   = 15 possibilites

They are

1,2      2,3      3,4    4,5    5,6

1,3      2,4      3,5    4,6

1,4      2,5      3,6

1,5      2,6

1,6

Only   1,6     2,5    and   3,4     add to  7

So....there are 12 ways to paint the faces so  that the numbers do not add to 7

Since arc SN  =  60°, the  angle 8  = 1/2 of this  = 30°

And arcs  SN, SQ, QP  and PN  sum  to 360°....so

SN   + SQ   + QP + PN

(60)  + ( x)  + (x + 40)  + (2x  - 16)  = 360      simplify

84  + 4x  = 360    subtract 84 from both sides

4x  = 276       divide both sides  by 4

x = 69°

So... arc QP  = x + 40  =  69 + 40  =  109

And angle 7  is 1/2 of this  = 54.5°

So  angle  6  = [ 180  - angle 8 - angle  7  ]  =   [ 180 - 30  - 54.5  ]  = 95.5°   

And since angle 5  is vertical to angle 6, its measure is also  95.5°

And angle  4 is supplemental to angle 6  so its measure is  180 - 95.5   = 84.5°

And angle 1  will have 1/2 the measure  of  the sum  of arcs SN and SQ =

(1/2)(60 + 69)   =

(1/2)(129)  =

64.5°

Because of the conjugate property, the function will also have the zero  -3 + 2i

So....let the  function be  f(x)  and we have that

f(x)  = a( x + 1)( x - ( - 3 - 2i) ) ( x - (-3 + 2i) )   simplify

f(x)  = a ( x + 1)  ( x^2  -x (-3 + 2i)  -x(-3 -2i)  + ( -3 + 2i) (-3 - 2i) )

f(x)  = a  ( x + 1) ( x^2 + 3x - 2xi + 3x + 2xi + 9 - 4i^2 )

f(x)  = a ( x + 1) (x^2 + 6x + 9 + 4)

f(x)  = a (x + 1)(x^2 + 6x + 13)

f(x)  = a  ( x^3 + 6x^2 + 13x  + x^2 + 6x + 13 )

f(x)  = a  ( x^3 + 7x^2 + 19x + 13)

And  since  f(-9)  = -6, we can find  "a"  as

-6  = a ( (-9)^3 + 7(-9)^2 + 19(-9) + 13)

-6  = a ( -320)

a  =  6 / 320   =  3/160

So.....f(x)  =   (3/160) ( x^3 + 7x^2 + 19x + 13)

∠BDE = ∠BAC

∠BED  = ∠BCA

AA Similarity Postulate

Segment Addition Postulate

I substituded into the formula, but I'm not sure if its correct. :(

HAHAHA!!!!...I wish  !!!!

But...THX anyway....LOL!!!!

We  know an angle and an adjacent side.....and we are  looking for the opposite side

The tangent  function relates all three....so we have...

tan (62°) = x / 40       multiply both sides by 40

40 * tan (62° )   =  x  ≈  75.2 yds

THE GOAT!!!!

We can use the intersecting chord theorem  that says that

CE * DE  = BE * AE        substituting, we have that

CE * 5 =  3 * 15

CE * 5   = 45        divide both sides by  5

CE  = 45/5  = 9 inches

Here's one thing  to remember....because the sine and  cosine are  co-functions ......

sin (90 - x)°   = cos x°  .....so letting x  =  19°, we have that

sin (90 - 19)°  = cos 19°

sin 71°  = cos 19°

So....x  = 71°

sin C  = h/a

substitution property of equality

associative  property  of  multiplication

Definition of Altitude

sin M  = h/p and sin P = h/m

Multiplication Property of  Equality

psin M     =  m sin P

_____         _________

  pm               pm

See #5   at:

https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html

If ADB  = 225°, then minor arc AB  = 360  - 225  = 135°

And the measure  of APB  =

(1/2) ( measure of arc ADB  -  measure of  minor arc AB )  =

(1/2) ( 225  - 135)  =

(1/2)(90)  =

45°

f(x)=3x+3

h(x)=e^(2x+1)

i(x)=x^2+4x+3

1.   r(x)=h(i(x)).....here. we are putting the function "i"  into the function "h"  

r(x)  = e^[(2 (x^2 + 4x + 3) + 1]

r(x)  = e^[2x^2 + 8x + 6 + 1]

r(x)  = e^[2x^2 + 8x + 7]

 2.   t(x)=f((h(i(x))).......here we are putting the first result into  "f"

t(x) = 3(e^[2x^2 + 8x + 7 ] ) + 3  =

t(x) = 3 [ e^(2x^2 + 8x + 7)  + 1 ]

sorry here is the one

https://learning.k12.com/d2l/common/viewFile.d2lfile/Database/Mzg1MzA1NQ/Quiz%204.11__4.JPG?ou=374067

Sorry, TM31....that is not a web address.....that is a file  address on the computer to which you are using/connected.

This is depressing.

solving this inequality?

\(-3x^2-12 \le 3x^2-5x-68\)

-3x^2-12 \le 3x^2-5x-68

\(\begin{array}{|rcll|} \hline -3x^2-12 &\le& 3x^2-5x-68 \\ -3x^2-12 -(3x^2-5x-68 ) &\le& 0 \\ -6x^2 + 5x + 56 &\le& 0 \\ \hline \end{array}\)

Now all we have to do is find the zeroes of \( y = -6x^2 + 5x + 56\)

\(\begin{array}{|rcll|} \hline -6x^2 + 5x + 56 &=& 0 \\ x &=& \frac{-5\pm \sqrt{25-4\cdot(-6)\cdot 56 } }{2\cdot(-6)} \\ x &=& \frac{-5\pm \sqrt{1369 } }{-12} \\ x &=& \frac{-5\pm 37 }{-12} \\\\ x_1 &=& \frac{-5+37}{-12} \\ x_1 &=& \frac{32}{-12} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-\frac{8}{3}} \\\\ x_2 &=& \frac{-5-37}{-12} \\ x_2 &=& \frac{-42}{-12} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{\frac{7}{2}} \\ \hline \end{array}\)

Since \(y = -6x^2 + 5x + 56\)  graphs as a down parabola,
the quadratic is below the axis on the ends:

Then the solution is:
    \(x \le -\frac{8}{3} \quad \text{ or } \quad x \ge \frac{7}{2}\)