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Thanks Heureka and guest,

I think my logic is similar to guests but I did not really understand guest's that well.

Here is my diagram.

1) On my diagram I have labeled the angles of the triangle ABC

2) Then using only the fact that corresponding angles on parallel lines are congruent, I have labelled all the relevant angles in black.

3) Now, using the fact that the angle sum of a triangle is always the same (180 degrees but that does not matter) I have filled in the 3rd angle on all of the triangles - they are in pink.

4) Now it can be seen that there are 4 similar triangles. (3 equal angles test)

  Also there are 3 parallelograms.  

5) Now the ratio of the areas of the 3 small similar triangles is                    4:9:16

6) so the ratio of corresponding sides of the 3 small similar triangles is       2:3:4

7) On the next diagram I will label 3 coresponding sides with their lengths   (in black)

8) Then I will use the fact that parallelograms have opposite sides of equal length to fill in the missint length secions on AC (in red)

9)   3+2+4=9

10)   So now I know that the ratio of the coresponding sides of the 4 similar triangles is    2:3:4:9

11) So the ratio of the areas of the 4 similar triangles is                                                      4:9:16:81

Since the areas of the little ones are exactly 4,9 and 16, the area of the big one, i.e. triangle ABC is 81 units squared.

3a  = 30  ⇒  a  = 10

3b = 15 ⇒ b = 5

3c = 24 ⇒  c = 8

So

c + b x a   =   8 + ( 5 x 10)  = 8 + 50   =  58

none of the numbers in your post showed up.

Here's  12

Assuming that  we are dealing with positive integers....let the odd numbers  be

2n - 1   and 2n + 1

Adding these together we get   4n

Sometimes.....in these kinds of proofs, it is good to see if we can discover a pattern....

Let n = 2

And the sum of the odds =  8

And the difference of two squares that equals this  is 3^2 - 1^2 = 9 - 1 = 8

We can write this as  [ 2n - 1 ]^2 - [ 2n - 3]^2

Note that the numbers in red are just  (n - 1) and (n + 1), respectively

Now......Let  n = 3

And the sum of the odds = 12

And the difference of two squares that equals this is 4^2 - 2^2  = 16 - 4  = 12

Which we can write as [ 2n - 2 ]^2 - [ 2n - 4 ]^2 

Note that, once again, the numbers in red are just ( n - 1)  and  ( n + 1), respectively

So....it appears that we can write the larger  square  as   [2n - (n- 1)]^2   =  [ n + 1]^2

And it appears that we can write the other square as  [ 2n - (n + 1) ]^2  =  [ n - 1]^2

And their difference is

[ n + 1]^2  -   [ n - 1]^2   =

[ n^2 + 2n + 1 ]  - [ n^2 - 2n + 1]  =

4n

5. Square ABCD has side length 60. An ellipse E is circumscribed about the square and there is a point P on the ellipse such that PC = PD =50. What is the area of E?

This one is a little difficult !!!!

Let the circle and the ellipse be centered at the origin

We can let the vertices of the square be  A = (-30,30) , B =(30,30), C = (30, -30)  and D = ( -30, -30)

The mid point of the bottom of the square is the point (-30, 0)

Call this point, M

And  PC = PD = 50

So....we can form right triangle MPC  such that PC forms the hypotenuse = 50

And MC = 30

So....MP  is the other leg  =  √ [ PC^2 - MC^2] = √[ 50^2 - 30^2]  = √ [ 2500 - 900 ] =

√1600 = 40

So......the distance from the origin to M  = 30

And the distance between M an P  =  40

So....using symmetry.....we can let the vertical axis of the ellipse  = 2* (30 + 40) = 140  = 2b

So......in the equation

x^2            y^2

___   +     ___    =  1

 a^2          b^2

We know that one point on the ellipse is (30, 30)  = (x , y)

And "b"  is (1/2) * 140  = 70

So.....we can find "a" as follows

30^2           30^2

_____  +    _____   =      1

  a^2            70^2

900            900

____ +      _____  =    1

a^2            4900

900         9

___  +   ____  =    1

a^2         49

900                   9

____  =   1  -    ___ 

 a^2                   49

900             [ 49 - 9]

___  =       _________

a^2                 49

900                 40

____   =        ____

 a^2                49

So

40 a^2 =  900*49

40a^2 = 44100

a^2  =  44100 / 40

a^2 = 4410/4

a = √ [ 4410 / 4 ]

And  the area of the ellipse   =

pi* a * b  =

pi * √[ 4410/ 4} * 70  ≈  7301.9   units^2

Here's the graph : https://www.desmos.com/calculator/r9mwhqlg2e

4. A circle centered at the origin intersects the ellipse  x^2/9 + y^2/16 = 1   at the four vertices of a square. Find the area of that square.

Since the diagonals of a square are at right angles and the center of the elliipse is at (0,,0).....the lines y = x  and y = -x    will form diagonals that intersect at (0,0) and will be at right angles to each other

To find out where  y = x intersects the ellipse.....let us solve this

[ sub x for y  in the ellipse equation ]

x^2/9 + x^2/16  = 1       multiply through by 144

16x^2 + 9x^2  = 144

25x^2  = 144

x^2   =  144 / 25    take both roots

x = ±√ [144/25]   = ±12/5   =  ±2.4

And siince  y = x     and y = -x the coordinates  of the top  vertices of the square are

(2.4, 2.4) , ( -2.4, 2.4)

The distance between the points  is 4.8  =  the side of the square 

So....the area of the square  is  side^2   = 4.8^2   = 23.04 units^2

Here's a graph : https://www.desmos.com/calculator/uld9mzxaet

3. Find the foci of the ellipse whose major axis has endpoints (0,0) and (13,0) and whose minor axis has length 12.

We have this equation

(x - h)^2            (y - k)^2

_______    +   ________   =      1

   a^2                  b^2

The center of the ellipse is   at  (6.5, 0)     =  ( h, k)

The major axis is along the x axis.....its length is 13

"a" is 1/2 of this 6.5 

The minor axis is along y

And b is 1/2 of this  =  6

The y coordinate of both foci lies on the x axis

And the foci are given by   (  ±√ [a^2 - b^2 ] + h , 0 )    =   ( ±√ [6.5^2 - 6^2] + 6.5 , 0)   =

( 2.5 + 6.5, 0 )     and   (-2.5 + 6.5)    =

(9, 0)    and  ( 4, 0)

Here's the graph : https://www.desmos.com/calculator/hxa6eulntu

I think it is only 10 ways.

Put the teal bead on first. It is the marker bead. This takes care of the rotation aspect.

Then you have 2 lots of 3 identical beads. so that is 6!/(3!3!) ways to place them. Which is 20 ways.

But the braclet can be flipped so I think this will halve the number of possibilities

For instance  TOOOBBB is the same as TBBBOOO

So I think the answer is 10 ways.

oof I guess I've uhh 'doomed' you all whoopsies...

1)\(5x + 8y = -9 <=> y=-\frac{5}{8}x-\frac{9}{8} \) 

λ1=-\(\frac{5 }{8}\) 

L2 is perpendicular to line L1 so λ1*λ2=-1 <=> -5/8*λ2=-1

λ2=\(\frac{8}{5}\)

so the l2: y-y1=λ2(x-x1)  x1=10,y1=10 because  passes through the point (10, 10)

so \(y-10 = \frac{8}{5}(x-10)<=> y=\frac{8}{5}x-\frac{80}{5}+10<=> y=\frac{8}{5}x-\frac{30}{5}\)

m=8/5,b=-30/5

m+b=\(\frac{8}{5}-\frac{30}{5}=\frac{-22}{5}\)

2)\(3x-7≤ 5x+9<=> 3x-5x≤9+7 <=> -2x≤16 <=> x≥ 8\)

\(x≥ 8\)

3)\(-2s+14>4s +8<=> 14-8 > 4s+2s <=> 6>6s <=> 1>s <=> s<1\)

\(s<1\)

 11) \(\frac{α}{β}> \frac{2α-β}{α+4β} \) multiply by α+4β and β and its >0 because α,β its positive numbers 

\(α(α+4β)>β(2α-β) <=> α^2+4αβ > 2αβ-β^2 <=> α^2 +4αβ - 2αβ + β^2 >0 <=> α^2 +2αβ + β^2 >0 <=> (α+β)^2>0 \)

where applicable! So we prove it!

12) Something is wrong here or I don't understand the question:

exp.:\(3^2-2^2=9-4=5\)  But there is no sum of two consecutive odd numbers be equal with 5 because 1+3 = 4 and general  sum of two consecutive odd numbers  is allways be equal with even number because if n is a number we have:

n+(n+2)=2n+2 = 2(n+1) for every n natural so we have 2(n+1) 

is divided by 2 so it's even number and 5(in this example) it's not even number.

 Hope this helps!

\(9x+17=71 <=> 9x=71-17 <=> 9x=54 <=> x=\frac{54}{9} <=> x=6\)

Hope this helps!

It has a negative leading coefficient becaus on the right hand side it ends down not up.

The degree must be even because the left side is down too   

It is not a parabola so there is probably a triple root at x=0. 

This is not a technical explanation it is just one that I hope you can learn from.

\(y=-x^3(x+2)\)

1. An ellipse has its center at the origin, its foci on the y-axis, and its major axis is three times as long as its minor axis. Given that the ellipse passes through the point (-4, 0), find its equation.

2. What is the minimum distance between a point on the circle x^2+y^2=16 and a point on the line x-y=8.

Thanks, Guest!!!

THAT is a REALLY nice method  !!!!

We have a SSA-type problem

Let the lighthouse be located at C......angle ACB  = 33°

     C

                36

A        20        B

We are looking for angle CBA

We can find angle CAB, first

Using the Law of Sines

sin ACB /20 = sinCAB / 36

sin (33) / 20  = sin(CAB) /36

(36/20)sin (33)  = sinCAB

arcsin [ (36/20)sin(33)] = CAB  ≈ 78.62°

Therefore  a possible value for  angle CBA   ≈ 180 - 33 - 78.62  ≈ 68.38°

Another possible value for angle CAB  =  180 - 78.62  ≈ 101.38°

Add  this to the known angle ACB  =  33 + 101.38   = 134.38°

Because this is < 180, we have two possible triangles

So....the other possible value for angle CBA  = 180 - 134.38  = 45.62°

Continuing on, we have \(\frac{24}{2}, \frac{-20}{2}\). Thus, our answer is \((12,-10)\).

Deleted.

Here is another way:

Triangle PD'D:PF'F and 9:16 and sqrt(9):sqrt(16)=3:4
Line PD':FP =3:4 and FD' =3 + 4=7. So, triangle PFF':FD'B.
FD'B =7^2 - 16 - 9 =24 area of PDF'B. By similar reasoning:
EF' =2 + 4 =6 and the area of EF'C=6^2 - 4 - 16=16 area of E'PFC. And the same applies to DE' =2 + 3 =5 and the area of ADE' =5^2 - 4 - 9=12 area of AEPD'.
So, the area of ABC triangle= 4 +9+16+24+16+12=81 units^2
 

This area will form a  right riangle

We can find one leg as follows

6  = (1/2)x     multiply both sides by 2

12  = x

The other leg  = 6

And the area of a right triangle =    product of the legs / 2   =

[ 12 * 6 ] /   2      =

72  / 2  =

36 units^2  

Here's a graph : https://www.desmos.com/calculator/0vjigygyh7

You can see two points on this line (0,1) and (2,0).  From this we can find the line.

\(\text{the slope }m \text{ is given by}\\ m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{0-1}{2-0} = -\dfrac 1 2\\ \text{using the first point and the slope we write the point slope form of the line}\\ (y-1) = -\dfrac 1 2 (x-0)\\ \text{and with a bit of algebra we get }\\ \dfrac 1 2 x +y= 1 \text{ or}\\ x+2y = 2\)

\(\text{Given two points }p_1,~p_2, \text{ the midpoint of the segment joining them is given by}\\ m=\dfrac{p_1+p_2}{2} \\ \text{in this case }\\ m =\left( \dfrac{-4+28}{2},~\dfrac{9-29}{2}\right)=?\)

I leave that bit of arithmetic to you.