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Simplifying the equation, we get \({x}^{2}+19x+5=0\). We know, that \(a+b=-19, ab=5\), by Vieta's formula.

Lets see if we can use this to our advantage here. A quadratic with roots \(\frac{1}{a+1}, \frac{1}{b+1}\) will have a sum of roots of \(\frac{1}{a+1} + \frac{1}{b+1} = \frac{b+1+a+1}{(a+1)(b+1)}=\frac{a+b+2}{ab+a+b+1}\). Plugging in the values we got from Vieta's, we get \(\frac{-19+2}{5-19+1}=\frac{-17}{-13}=\frac{17}{13}\). 

A quadratic with roots \(\frac{1}{a+1},\frac{1}{b+1}\), will have a product of roots of \((\frac{1}{a+1})(\frac{1}{b+1})=\frac{1}{ab+a+b+1}\). Again, plugging in the values from Vieta's, we get \(\frac{1}{5-19+1}=-\frac{1}{13}\). 

If you're looking for a monic quadratic, a possible answer would be \({x}^{2}-\frac{17}{13}x-\frac{1}{13}\). An equalivent, non monic quadratic is \(13{x}^{2}-17x-1\). Note, there are infinitely many solutions because of different possibilites of leading coefficients.

The radius of the circle can be  found as

25 pi = pi *r^2

25 = r^2

r = 5

Ler R be the radius of the sphere

R^2 =   [ 5^2 + r^2 ] =  [ 5^2 + 5^2 ]  =  50

Surface area of sphere  =

4 pi R^2  = 

4 pi * 50  =

200 pi

The volume of the resulting shape, is the area of the cube - 8*the area of each pyramid. (8 corners).

The volume of the cube, is 6*6*6=216.

We see that each pyramid is made out of 3 right isoceles triangles, and one equaliteral. We see that each right isoceles triangle has a leg length of 3, because 2 leg lengths make 6.

To find the volume of each pyramid, we use the formula, \(Volume = \frac{1}{3}base*height\), but we should choose carefully what we want our base and height to be. If we let the equilateral triangle be our base, and the distance from that triangle to the corner be our height, that is some really hard calculations. Instead, we see that each pyramid corner that is cut off is made from 3 right isosceles sides, with one equilateral triangle triangle. If we let an right isoceles triangle be our base, we get much easier calculations.

The area of the base is. \(\frac{3*3}{2}=\frac{9}{2}\), and using the volume formula, each pyramid has an area of \(\frac{1}{3}*\frac{9}{2}*3=\frac{9}{2}\). So the area of the resulting polyhedron is \(216-\frac{9}{2}*8=180\).

What do we know about tan(x) and sec(x)? We know \(\tan(x)=\frac{\sin(x)}{\cos(x)}, sec(x)=\frac{1}{cos(x)} \). Now use these, and replace them in the equation. 

We get \(\frac{sin(x)}{\cos(x)}+\frac{1}{cos(x)}=0\). Multiplying by cos(x), with the restriction, cos(x) is not equal to 0, gives us \(\sin(x)+1=0\), or \(sin(x)=-1\). So, now we use the formula, \({\sin}^{2}(x)+{\cos}^{2}(x)=1\), we get \(cos^2(x)=0, cos(x)=0\). Notice there is only one solution, because +/- 0 is always 0.

But we see from ealier that cos(x) is not 0, so we have no solution.\(\)

Hint: make the small length of the shaded area, ( the smaller side of the parallelogram x). Then you can use simular triangles.

Our current goal is to find the value of xy, so we can get x^2 + y^2 = (x + y)^2 - 2xy.

x^3 + y^3 can be factored to (x + y)(x^2 - xy + y^2)

Additionally, x^2 - xy + y^2 = (x + y)^2 - 3xy 

So, x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)[(x + y)^2 - 3xy], now substitute:

2 * (4 - 3xy) = 5  => 4 - 3xy = 2.5 => 1.5 = 3xy => xy = 0.5

Now, to get x^2 + y^2, just do (x + y)^2 - 2xy = 4 - 1 = 3

\(\angle TOP=\angle UOP=45^{\circ}\\ \angle TOU=\angle TOP+\angle UOP=2\cdot 45^{\circ}\\ \color{blue}\angle TOU=90^\circ\)

!

If one side is  12, the sum of the lengths of the other two sides must be > 12

The smallest possible perimeter  is  > 24

RT / TS =  2/5

So  TS  = ( 5) / (2 + 5) SR=  (5/7)SR = (5/7)QP

Triangle SUT similar to triangle QUP

So height of  SUP  = (5/7) height of QUP

So VR  = (5/7)QV

So  QR  =  QV + VR  =  QV + (5/7)QV

So  QR  =  (12/7)QV

So (7/12)QR = QV

So  QV / QR  = 7/12 

And triangles QVU and QRS are similar

So 

UV / QV  = SR / QR

UV  =  QV * SR / QR  =   

UV  = (QV / QR) * SR  =  (7/12) (20) =  140  / 12 =  35 / 3 =  11 + 2/3 

Radius of semi-circle  =  (9 + 16 + 9) / 2  = 34 / 2  =  17

Height of trap  =   sqrt [    17^2 - (QP/2)^2 =  sqrt [ 289  - (16/2)^2] = sqrt [ 289 - 64 ] = 25

Area of  trap  =   (1/2) (25) ( 34 + 16)  =  25 * 25  =  625

P is not defined.....

[ LOM ] =  (1/2) r^2 sin (90°)  =   (1/2)(1)^2 * (1)  =  1/2

We see that triangles EAD~EBF by AA similarity.

But we also know that they have the same area, so they must be congruent.

So, AE = BE = 1/2AB = 1/2DC.

The area of triangle FDC is \(4*\frac{{DC}^{2}}{{EB}^{2}}\), or 4 times the square of the ratio of the bases. And because DC = 2EB, the area of FDC is 4*4=16.

The area of ABCD is 16-4+4 = 16.

(a + 2) / ( a + 5)  =  b /4

4 (a + 2)  = b ( a + 5)

4a  + 8 = ab + 5b

b ( a + 5)  = 4a + 8

b =  ( 4a + 8) / (a + 5)

b =  4 ( a + 2)  / (a + 5)

b will never be less than 5 when  a approaches negative infinity

b will never be more than 3 when a approaches positive infinity

a         b

-17     5

-11     6

-8       8

-7       10

-6       16   

-4       -8

-3       -2

-2        0

-1        1

 1        2

 7        3

Here :   https://web2.0calc.com/questions/square-and-triangles#r1

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Here :  https://web2.0calc.com/questions/sector_4

EF  parallel to QB

So triangle DEF is similar to  triangle DQB

 F is the midpoint of DB (the diagonal of the square)

So

DF / DB   =    EF / QB

(1/2)DB  / DB  =  EF / QB

(1/2) = EF / 1

EF =1/2

And EF is the diagonal of the square formed by the overlap

So.....the side of the squre  =   (1/2)(1/sqrt2)  =  1/ sqrt 8

Area of the quadrialeral   =  (1/sqrt8)^2  =   1/8

n = 2

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