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The general formula across line y = x, is \((a, b)\rightarrow (b, a)\), so the result is \((7, 5)\).

But if you want a more general solution that can work for any line, then it is much more difficult, but turning to imaginary numbers + roots of unity is one way to do it.

Set the side lengths of the triangle to be \(x, x, y\)

\(2x+y=5\), so \(2x<5\) and \(x\le2\).

Therefore, because x is a positive integer, x is either 1 or 2.

But we need to be careful because 1, 1, 3 is not a triangle.

So the only possibility is 2, 2, 1.

AC^2 + [(1/2)BC]^2  = AM^2

[(1/2)AC]^2 + BC^2 = BN^2

AC^2 + BC^2/4 =  49

AC^2/4 + BC^2   =  36

Multiply the first equation through by 4   and the  second through by -1

4AC^2 + BC^2  = 196

-AC^2/4-BC^2  = - 36         add these

(15/4)AC^2  =  160

AC^2   =  4 * 160  /  15

AC^2   = 128/3

To find BC

4 (128/3) + BC^2  =  196

(512/3) + BC^2 = 196

BC^2  =  196 - 512/3

BC^2  = 76/3

AC^2 + BC^2  =  AB^2

128/3 + 76/3 = AB^2

204/3  = AB^2

68 = AB^2

AB = sqrt (68)  =  2sqrt (17)  =  hypotenuse length

The coefficient on x comes  from

x (14)  + 1 (-23x)   =

14x  - 23x   =

-9x

Center = (m,n)  =   (0,0)

a= 2

b = 2

The hyperbola intersects the y axis.....so the form is

(y  - m)^2 /  a^2    -  ( x  - n) / b^2  =   1

(y - 0 )^2  / 4  -  ( x - 0 )^2   =  1

y^2  / 4   -  x^2 / 4   =   1

Since AM is a median, then M is the midpoint of BC. Therefore, BM=MC=6.

Similarly, since BN is a median, then N is the midpoint of AC. Therefore, AN=NC=7.

By the Pythagorean Theorem on right triangle ABM, AB2=AM2+BM2=72+62=85. Then, AB=85​.

Similarly, by the Pythagorean Theorem on right triangle ACN, AC2=AN2+NC2=72+72=98. Then, AC=98​.

Finally, by the Pythagorean Theorem on right triangle ABC, BC^2 = AB^2 + AC^2 = 85 + 98 = 183​.

Therefore, the length of the hypotenuse is BC = sqrt(183).

BP = 22/3

r^2 = 624

Both solved with CAD, probably not how you're supposed to be doing it but I think it's right

Here are the answers to your question:

e^(pi*i/2) = 1/2 + 1/2*i.

e^(-2*pi*i/3) = 1/sqrt(2) - 1/sqrt(3)*i

e^(9*pi*i/4) = 1/3 + sqrt(2)/4*i

Hope this helps!

First one

BA (BA + BS)  = AC ( AC + PC)

3 ( 3 + 21)  = 8 ( 8 + PC)

72  = 8 (8 + PC)      divide both sides by 8

9  = 8 + PC

PC  = 1

Pythagorean Theorem

AS = 24

AP = 9

sqrt (AS^2 - AP^2)  = SP  =  sqrt [ 24^2 - 9^2]  = sqrt (495)

Pythagorean Theorem again

SC =  sqrt ( SP^2 + PC^2 )  = sqrt ( 495 + 1)  = sqrt (496)  =   4sqrt (31)

Because SPC is a right angle,  then  SC is the diameter of the circle

Then the radius is  2sqrt (31)

Area of circle  = pi * (2sqrt (31))^2  =  124 pi

When Edith finishes the race, how far behind is Foley, in km?

Hello Lilliam0216!

\(v_d=24km/t_d\\ v_e=(24-5)km/t_d=19km/t_d\\ v_f=(24-6)km/t_d=18km/t_d\\ t_e=\frac{24km}{v_e}=\frac{24km\cdot t_d}{19km}\\ s_f=v_f\cdot t_e=\frac{18km}{t_d}\cdot \frac{24km\cdot t_d}{19km}=22.737km\)

Foley was (24-22.737)km=1.263km behind when Edith finished the race.

  !

(x^2 - 2x - 5) f(x) = 2x^4 + 19x^3 + kx^2 - 15x - 1, what does this mean?

Apply the heron's formula directly on triangle PQR:

The semiperimeter is \(\frac{36+22+26}{2}=42\).

\([PQR] = \sqrt{42*(6)*(20)*(16)}=\sqrt{80640}=48\sqrt{35}\).

(The [PQR] notation means area).

x^3 ( x^2 + tx)  =  x^5 + tx^4.....it doesn't have an x^2 term anyway

x^2 + 23x  + k

To have a double root, the discriminant  must =  0 

23^2  - 4 *1 * k   =  0

529 =  4k

k = 529 / 4

16r^3 = 26

r^3  = 26/16  = 13/8

r = (13/8)^(1/3)

24th term =  16 (13/8)^(1/3)  ≈  18.81

x^2 + 8x +  p

p  = (1/2 the coefficient on  x )^2  =  (8/2)^2  =  4^2 =  16

So

x^2 + 8x + 16  =

(x + 4) ( x + 4)  =

(x + 4)^2

We want the form of the ellipse to be    x^2 / a^2  + y^2 / b^2  = 1

x^2 + 2y^2  =  32       divide through by 32

x^2 / 32  +  y^2 / 16  =   1

b^2   = 16

b   =  4 =  semi-minor axis   =   the   radius of the  circle....so r^2 =  16

x^2 + y^2  = 16 

Simplify as

19x^2- 6x + 31  =  0

x = [   6 +/- sqrt [ 36 - 4*19*31] ] / 38  =  [ 6 +/- sqrt [-2320] ]/ 38  = 

[ 6 +/- 4 i sqrt [ 145] ]  / 38

a = 6/38  = 3/19

b = 4sqrt [145 ] / 38  =  2sqrt [ 145] /19

a * b  =    (3/19)  (2sqrt [145] / 19 )  =  6sqrt[145] / 361

The reason for the loss of points and missing posts is on or about February 29 or March 1, 2024 the web2.0calc databases (and maybe the operating systems) were restored to January 16, 2024. The German forum was restored to December 8, 2023.  All posts (and related point tallies) subsequent to the restore point and prior to the restore time no longer exist on the forum.

I can only speculate on the reasons for restoring to a point six weeks prior: A corrupted database index is the most likely cause. These can be difficult and time consuming to repair if the errors are nonlinear or compounded over a large span of time.   

I wonder if it is due to the amount of repeat questions (and most likely homework questions) being posted in this time period? Anyone have any answers???

The vast majority of questions posted on the forum are posted by scripted bots, under the control of a few psychopathic and obsessed trolls. Some of the answers are also posted by scripted bots; these answers may be generated by AI, and at least 20% are incorrect, with wrong or missing procedures and/or intentional math errors. There are a few legitimate questions posted by humans. There are also bot-posted questions that imitate   

This forum is a real microcosm of the science fiction narratives where the world is invaded and takeover by human-like cybernetic virtual beings under the control of biological humans. It’s bloody creepy!!!

GA

--. .- 

Multiply through by  (x + 5) (2x + 8)

(x + 4)(2x + 8)  = (x - 3)(x + 5)  - 7(x+5)(2x + 8)

2x^2 + 16x + 32  =  x^2 + 2x -15  - 7 (2x^2 + 18x + 40)

2x^2 + 16x + 32 = x^2 + 2x - 15 - 14x^2 - 126x - 280

15x^2 + 140x + 327  =  0

The solutions  are

[-140  +/- sqrt [ 140^2 - 4*15*327] ] / 30    =   [-140 +/- sqrt [ -20 ] ] / 30  =

[ -140 +/- 2isqrt (5) ] / 30   =

-14/3  +/- (sqrt (5)  / 15)  i

Simplfy  as

-8x^2 -kx + 47   = 0

We have no real solutions when the discriminant is <  0

k^2 - 4 (47)(-8)  <  0

k^2 + 1504 < 0

The left side is always  positive for any k.....so.......we will have real solutions for any k

80  = (1/2) (12) ( b + b + 4)

80 / 6  =   2b + 4

40/3  = 2b + 4         multiply through by 1/2

20/3  = b + 2

b = 20/3 - 2   = 20/3 - 6/3 =    14/3

b + 4 =  14/3 + 12/3  =   26/3

Assuming that we have an  isosceles trapezoid, we  can  find the length of a side by constructing a right triangle with legs of   12  and  [ (b + 4) - b ] / 2  = 2

The side length is  sqrt [ 12^2 + 2^2]  = sqrt (148) = 2sqrt (37)

The perimeter  is    14/3 + 26/3 + 2 * 2sqrt (37)  =  40/3 + 4sqrt (37)  ≈  37.66