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f(2)  = 2^2 + 2sqrt (2)  = 4 + 2sqrt (2)

2f(2)  = 8 + 4sqrt (2)

f(8)  = 8^2 + 2sqrt (8) =     64  + 2*2sqrt (2)  =  64 + 4sqrt (2)

2f(2) - f(8)  =

(8 + 4sqrt (2)  )  -  ( 64 + 4sqrt(2) )  =    - 56

Hey! That's not nice. I still need help!

The common  ratio  =  (2/3) / (3/2)  =  (2/3) (2/3)  =   4/9

Sum  =

(3/2)  / ( 1 - 4/9)  =

(3/2) / ( 5/9)  =   (3/2) ( 9/5)    =   27  / 10

Nevermind, I solved it.  But I need help with

https://web2.0calc.com/questions/in-triangle-abc-let-the-angle-bisectors-be-by-and-cz

https://web2.0calc.com/questions/in-triangle-abc-let-the-angle-bisectors-be-by-and-cz_1

Here's Geno's answer  : https://web2.0calc.com/questions/find-all-pairs-x-y-10-of-real-numbers-such-that-x

The solutions are

( 5 + sqrt 3 , 5-sqrt 3)  and (5 -sqrt 3  , 5 + sqrt 3)

thank you!

Circumradius =  product of the sides  /   (4 * area)

Height =  sqrt [ 29^2 - 20^2 ] =  sqrt [ 441] =  21

Area  = (1/2) (40) ( 21)  =  420

Circumradius  =      29^2 * 40  / (4* 420)  =  841/42  ≈  20.02

s/2 = 3 - 5t   mult through by 2 →  s =  6  -10t

3t - 6(6 - 10t)  = 9 - 5t

3t -  36 + 60t  = 9 - 5t

63t + 5t =  9 + 36

68t =  45

t  =  45 / 68

s = 6 - 10 (45/68)   =   -21 / 34

Draw BO  and CO

[ABO ]  = (1/8)  of the octagon's area

[BCO ] =  (1/8) of the octagon's area

And 

[BMO ]  = (1/2) [ BCO ] =   1/16  of the octagon's area

And  [ ABO ] + [ BMO ]  =   [ ABMO ]  =   (1/8) + (1/16) =  (3/16) of the octagon's area

So

[ABMO ] / [ ABCDEFGH]  =  3 / 16

See the graph here : https://www.desmos.com/calculator/doygww4j6l

c =   .844

a  = -11.844

c  - a  =   .844  -  ( - 11.844 )  =   .844 + 11.844    ≈  12.688   

-4.9t^2  + 42t + 18.9  = 84

This might be best solved with a graph (in varable x)  here : https://www.desmos.com/calculator/agj3c43ypd

It first reaches 84 m  at   t  ≈   2.03  sec

3x  + y  +  xy   = 115

y =  (115 - 3x)   / ( 1 + x)

When x = 1   y =  56

So

x + y =  57

The sum of the interior angles = 360

So

(128 - 3d) + (128 - 2d ) + (128 - d) + 128  =   360

4 (128) - 6d  = 360

4 (128) - 360  =  6d

152  =  6d

d = 152 / 6  =  76/3 

The smallest angle =    (128 - 3(76/3) )  =  128 - 76  =   52°   (as we might have guessed....it should be supplemental to 128°)

2(x- 8)^2  + 4   =

2 (x^2 -16x + 64) + 4

2x^2 - 32x + 128 + 4

2x^2 - 32x  + 132                   a = 2    b  =  -32          c = 132

So

3ax^2  + 2bx + c   =

8x^2 - 64x + 132  =

8 ( x^2 - 8x )  + 132             complete the square on  x

8 ( x^2 - 8x + 16) + 132 - 8(16)

8 ( x - 4)^2  +  4

h =  4 

x + 2y =  8         square both sides

x^2  + 4xy  + 4y^2  =   64

x^2 + 4(8) + 4y^2  =    64

x^2 + 4y^2  =  64 - 4(8)

x^2 + 4y^2  =  32

uh srsly get yur own thread.

This will  evaluate  to    

            1

________________________  =  

x * x^2  * x^3 * .........x^9*x^10

                1

_________________________     = 

        x^ (1 + 2 + 3 +... +9 + 10)

                    1

____________________   = 

          x^[ 10 * 11  / 2 ]

                     1

              ________

                   x^55

So

                 1

        __________

               2^55

And help with these.  These need to be done soon.

11) In the right triangle, an altitude is drawn from the right angle to the hypotenuse. Circles are inscribed within each of the smaller triangles.  What is the distance between the centers of these circles?

12) An equilateral triangle and a square both have perimeters of 48 inches. What is the ratio of the area of the triangle to the area of the square? Express your answer as a common fraction.

13) The square below has area 100. Find the area of the shaded region

14) In triangle ABC, points D and F are on AB and E is on AC such that DE || BC and EF || CD. If AF = 9 and DF = 2, then what is BD? 

15) Points A B and C are on a sphere whose radius is 13. If AB = BC = 12, what is the longest possible value of AC?

um, u know you can post your own questions right?

Need help with these too:

5) In the trapezoid shown, the ratio of the area of triangle ABC to the area of triangle ADC is 7:2. If AB+CD=210 cm, how long is segment AB?

6)  Triangle PAB is formed by three tangents to circle O and angle PAB = 45 . Find angle AOB.

7) In the circle with center Q, radii AQ and BQ form a right angle. The two smaller regions are tangent semicircles, as shown. The radius of the circle with center Q is 15 inches. What is the radius of the smaller semicircle? Express your answer as a common fraction. 

8) In a circle, the length of an arc intercepted by a central angle is 12mm, and the radius of the circle is 8mm. What is the measure, in radians, of the angle?

9) Three balls are placed inside a cone such that each ball is in contact with the edge of the cone and the next ball. If the radii of the balls are 20 cm, 12 cm, and r cm from top to bottom, what is the value of r?

10) Right triangle ABC has side lenghts AB = 3, BC = 5. Square XYZW is inscribed in triangle ABC with X and Y on line AB, and Z on line BC. What is the side length of the square.

We will multiply 1000000  by (5/2) four times (to get the 9th term) and then multiply this result by (1/2) to get the tenth term...so....

1000000 (5/2)^4 * (1/2)  =

500000 * (5/2)^4

500000 * (2.5)^4    =      19,531,250

We want to compute \(\cos \frac{k(k - 1) \pi}{2}\) for various intergers k. The first values are as follows:

 \(\begin{array}{c|c} k & \cos \frac{k(k - 1) \pi}{2} \\ \hline 1 & 1 \\ 2 & -1 \\ 3 & -1 \\ 4 & 1 \\ 5 & 1 \\ 6 & -1 \\ 7 & -1 \\ 8 & 1 \end{array}\)

The values seem to be periodic with period 4. If we let \(f(k) = \frac{k(k - 1) \pi}{2},\) then 

\(f(k + 4) - f(k) = \frac{(k + 4)(k + 3) \pi}{2} - \frac{k(k - 1) \pi}{2} = (4k + 6) \pi \\ = (2k + 3) \cdot 2 \pi\)

Since f(k+4) and f(k) differ by a mutiple of \(2 \pi\), \(\cos f(k)\) is indeed period with period of 4. So for a positive integer m, 

\(A_{4m - 1} = \frac{(4m - 1)(4m - 2)}{2} \cos f(4m - 1) = -\frac{(4m - 1)(4m - 2)}{2}, \\ A_{4m} = \frac{(4m)(4m - 1)}{2} \cos f(4m) = \frac{(4m)(4m - 1)}{2}, \\ A_{4m + 1} = \frac{(4m + 1)(4m)}{2} \cos f(4m + 1) = \frac{(4m + 1)(4m)}{2}, \\ A_{4m + 2} = \frac{(4m + 2)(4m + 1)}{2} \cos f(4m + 2) = -\frac{(4m + 2)(4m + 1)}{2},\)

Therefore:

\(A_{4m - 1} + A_{4m} + A_{4m + 1} + A_{4m + 2} = -\frac{(4m - 1)(4m - 2)}{2} + \frac{(4m)(4m - 1)}{2} + \frac{(4m + 1)(4m)}{2} - \frac{(4m + 2)(4m + 1)}{2} = -2.\)

Summing over \(5 \le m \le 24,\) we find that \(A_{19} + A_{20} + \dots + A_{98} = 20(-2) = \boxed{-40}.\)

You should try these problems too!

11) In the right triangle, an altitude is drawn from the right angle to the hypotenuse. Circles are inscribed within each of the smaller triangles.  What is the distance between the centers of these circles?

12) An equilateral triangle and a square both have perimeters of 48 inches. What is the ratio of the area of the triangle to the area of the square? Express your answer as a common fraction.

13) The square below has area 100. Find the area of the shaded region

14) In triangle ABC, points D and F are on AB and E is on AC such that DE || BC and EF || CD. If AF = 9 and DF = 2, then what is BD? 

15) Points A B and C are on a sphere whose radius is 13. If AB = BC = 12, what is the longest possible value of AC?

Here are some other geometry problems to try!

5) In the trapezoid shown, the ratio of the area of triangle ABC to the area of triangle ADC is 7:2. If AB+CD=210 cm, how long is segment AB?

6)  Triangle PAB is formed by three tangents to circle O and angle PAB = 45 . Find angle AOB.

7) In the circle with center Q, radii AQ and BQ form a right angle. The two smaller regions are tangent semicircles, as shown. The radius of the circle with center Q is 15 inches. What is the radius of the smaller semicircle? Express your answer as a common fraction. 

8) In a circle, the length of an arc intercepted by a central angle is 12mm, and the radius of the circle is 8mm. What is the measure, in radians, of the angle?

9) Three balls are placed inside a cone such that each ball is in contact with the edge of the cone and the next ball. If the radii of the balls are 20 cm, 12 cm, and r cm from top to bottom, what is the value of r?

10) Right triangle ABC has side lenghts AB = 3, BC = 5. Square XYZW is inscribed in triangle ABC with X and Y on line AB, and Z on line BC. What is the side length of the square.

4331  + 1052  +  ( 459 - N /2)  = 5543

5383  + 459  - N /2  =  5543

5842 - N /2  =  5543

-N / 2  = 5543 - 5842

-N/ 2 =  -299

N/ 2  = 299

N = 299 (2)  =   598