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o c**p thats so kind of you

thank u so much!

 9 x + I<3 (3 x + u) [Is I supposed to be the imaginary i??. If so, then inequalities are not well-defined in the complex plane (the field of complex numbers is not totally-ordered).

9x + I < 9x + 3u  subtract 9x from both sides

I < 3u                    and u element R.

convert 5.94x10to the 7th power =59,400,000

2.104x10to the -8 power =0.00000002104

1.2x10 to the power 5=120,000

7.2x10 to the -4 =0.00072

2.75x10to the 8th power =275,000,000

6.0502x10to the -3 power= 0.0060502

ERROR: in line 1: too few arguments for function between(float x, float y).

LOL recently I am addicted in writing C programs... 

Well. I got quick ways for all of them.

Formula: \(\displaystyle\int^{1}_{0}x^p(1-x^r)^s\; dx = \dfrac{s}{p + rs + 1}B(\dfrac{p+1}{r},s)\)

Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}\)

1)

 \(\displaystyle \int^{1}_{0}\sqrt{1-\sqrt{x}}dx\\ =\displaystyle \int^{1}_{0}x^{0}(1-x^{1/2})^{1/2}\; dx \\ =\dfrac{1/2}{0+1/2\cdot1/2 + 1}B(\dfrac{0+1}{1/2},1/2)\\ =\dfrac{2}{5}B(2,1/2)\\ =\dfrac{2}{5}\dfrac{\Gamma(2)\cdot\Gamma(1/2)}{\Gamma(2+1/2)}\\ =\dfrac{8}{15}\)

2)

\(\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}-\sqrt{x}}\; dx\\ =\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}(1-\sqrt[6]{x})}\;dx\\ =\displaystyle\int^{1}_{0}x^{1/6}(1-x^{1/6})^{1/2}\;dx\\ =\dfrac{1/2}{1/6+1+1/6\cdot1/2}B(\dfrac{1/6+1}{1/6},1/2)\;dx\\ =\dfrac{2}{5}B(7,1/2)\\ =\dfrac{4096}{15015}\)

3)

\(\displaystyle\int^{1}_{0}(x^{2/5}-\sqrt[3]{x})^{4/7}dx\\ =\displaystyle\int^{1}_{0}(x^{2/5})^{4/7}(1-x^{-1/15})^{4/7}dx\\ =\displaystyle\int^{1}_{0}x^{8/35}(1-x^{-1/15})^{4/7}dx\\ =\dfrac{\frac{4}{7}}{\frac{8}{35}+1+\frac{-1}{15}\cdot\frac{4}{7}}B(\dfrac{\frac{8}{35}+1}{\frac{-1}{15}},\dfrac{4}{7})\\ =\dfrac{12}{25}B(-\dfrac{129}{7},\dfrac{4}{7})\\ \approx -0.0626\)

4)

\(\displaystyle\int^{\pi/2}_{0}\cos^{15}xdx\\ =\displaystyle\int^{\pi/2}_{0}\cos x (1 - \sin^2 x)^7dx\\ u = \sin x\\ =\displaystyle\int^{1}_{0}(1-u^2)^7du\\ =\dfrac{7}{0+1+2\times7}B(\dfrac{0+1}{2},7)\\ =\dfrac{7}{15}B(1/2,7)\\ =\dfrac{2048}{6435}\)

5)

\(\displaystyle\int^{\pi/2}_{0}\sin^{13}xdx\\ =\displaystyle\int^{\pi/2}_{0}(-\sin x)(1-\cos^2x)^6\;dx\\ u = \cos x\\ =\displaystyle\int^{1}_{0}(1-u^2)^6du\\ =\dfrac{6}{0+1+2\times 6}B(\dfrac{0+1}{2},6)\\ =\dfrac{6}{13}B(\dfrac{1}{2},6)\\ =\dfrac{1024}{3003}\)

The only way the minor (or major) axis of an ellipse can subtend an angle of 90° at the focii is if the ellipse is a circle! This has an eccentricity of 0. 

Polynomials

Find a, b and the other factor

\(\text{factor } (x+1) \Rightarrow \text{let } x_1 = -1 \\ \text{factor } (x+2) \Rightarrow \text{let } x_2 = -2\)

Vieta's formulas:

\(\begin{array}{|rcll|} \hline -8 &=& -(x_1x_2x_3) \\ 8 &=& x_1x_2x_3 \\ 8 &=& (-1)(-2)x_3 \\ 8 &=& 2x_3 \\ \mathbf{x_3} &\mathbf{=}& \mathbf{4} \\ x_3 = 4 \Rightarrow \text{factor } (x-4) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline -a &=& -(x_1+x_2+x_3) \\ &=& (-1)+(-2)+4 \\ &=& -3+4 \\ \mathbf{a} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline -b &=& x_1x_2+x_1x_3+x_2x_3 \\ b &=& -(x_1x_2+x_1x_3+x_2x_3) \\ &=& -\Big((-1)(-2)+(-1)4+(-2)4)\Big) \\ &=& - (2-4-8) \\ &=& - (2-12) \\ &=& - (-10) \\ \mathbf{b} &\mathbf{=}& \mathbf{10} \\ \hline \end{array}\)

 4x^2+ 5x +2    factor out the 4

4 ( x^2  +  (5/4)x  + 2/4)

Take  (1/2)  of (5/4)  = 5/8......square it  = 25/64...add it and subtract it

4 ( x^2 + (5/4)x + 25/64  +  2/4 -  25/64 )

Factor the first three terms in the parentheses and simplify

4 [ ( x + 5/8)^2  +  32/64  - 25/64 ]

4 [ ( x + 5/8)^3  +  7 / 64]

4 ( x + 5/8)^2 +  7/16

First let's find the  x  values that make it equal  0 .

-5x² + 17x - 12  =  0          Using the quadratic formula....

x  =  [  -17 ± √( 17² - 4(-5)(-12) )  ] / [  2(-5)  ]  =  [  -17 ± 7  ] / [  -10  ]

x  =  1     or     x  =  12/5

The graph of this expression is a parabola that opens down, so

the  x  values that make it less than  0  will be in the interval

(-∞, 1) U (12/5 , ∞)

To make sure, let's test points.

-5(0)² + 17(0) - 12  =  -12  <  0

-5(2)² + 17(2) - 12  =    2   >  0

-5(3)² + 17(3) - 12  =  -21  <  0

Find the solutions x of the equation 2ix^2 + x +3i = 0

\(\begin{array}{|rcll|} \hline \mathbf{ax^2+bx+c} &\mathbf{=}& \mathbf{0} \\ \mathbf{x} &\mathbf{=}& \mathbf{{-b \pm \sqrt{b^2-4ac} \over 2a}} \\\\ 2ix^2 + x +3i &=& 0 \quad & | \quad a = 2i \quad b = 1 \quad c = 3i \\ x &=& \dfrac{-1\pm \sqrt{1^2-4\cdot(2i) \cdot (3i)} } {2\cdot 2i } \\ &=& \dfrac{-1\pm \sqrt{1-24i^2} } {4i } \quad & | \quad i^2 = -1 \\ &=& \dfrac{-1\pm \sqrt{1+24} } {4i } \\ &=& \dfrac{-1\pm \sqrt{25} } {4i } \\ &=& \dfrac{-1\pm 5 } {4i } \cdot \dfrac{i}{i} \\ &=& \dfrac{ (-1\pm 5)i } {4i^2 } \quad & | \quad i^2 = -1 \\ &=& \dfrac{ (-1\pm 5)i } {-4} \\ \\ x_1 &=& \dfrac{ (-1+ 5)i } {-4} \\ &=& \dfrac{ 4i } {-4} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{-i} \\\\ x_2 &=& \dfrac{ (-1- 5)i } {-4} \\ &=& \dfrac{ -6i } {-4} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{ \dfrac{3} {2}i } \\ \hline \end{array}\)

y = -5x^2+17x-12 

We can find the line of symmetry by finding the x coordinate of the vertex....this is given by

-17 / (2 * -5)  = -17/-10 =  17/10

So...the equation for the line of symmetry is just   x  = 17/10

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

If ax^2-5x+9=0  has only one root, the discriminant must = 0

Therefore

(-5)^2 - 4 (a)(9)  = 0

25  = 36a

a = 25/36

Here's a graph proving this :  https://www.desmos.com/calculator/puzyhvuy39

Are you asking about this part:

 (  [  (m + 2n) ]   - [m - 2n] - [2m]  )  / [  2 (m + 2n) (m - 2n) ]          ???

Starting from the previous step...

\(\frac{1}{2(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{m}{(m+2n)(m-2n)} \)

                                                                                               Multiply the third fraction by  2/2  .

\(=\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{2m}{2(m+2n)(m-2n)} \\~\\ =\, \frac{(m+2n)-(m-2n)-(2m)}{2(m+2n)(m-2n)}\)

Let R  = No. Red, B  = No. Blue, W = No. White , G  =  No. Green  and T  = Total

And we have that 

R / T  *  (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

B/T  *  ( R ) /[ T - 1] * (R - 1) /  [T - 2]  * (R - 2) /  [ T --3]   ⇒

R * (R -1) * (R - 2) * (R - 3)  =  B * R * (R - 1) * (R -2) ⇒

(R - 3)  =   B        

And

R / T  *  (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

W/T  * (R -3)/ [T -1] * R/ [ T - 2] * (R -1) / [T - 3] ⇒

R (R -1) (R - 2) ( R - 3)  = W * (R -3) (R) ( R -1)  ⇒

(R  - 2)  =  W 

And

R / T  *  (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

R /T *  (R -3)/[ T - 1] * (R -2) / [ T - 2] * G / [T - 3]  ⇒

R  (R -1) (R - 2) (R -3)  =  R (R -3) (R -2) G ⇒

(R -1)  = G

So  

R  = R

B  = R - 3

W = R - 2

G = R - 1

Let the number of blue = 1

Let the number of white  = 2

Let the number of green  = 3

Let the number of red = 4

Probability  of drawing 4 red  =     

[ 4 * 3 * 2 * 1 ]  / [10 * 9 * 8 * 7 ] = 24 / 5040

Probability of   one blue, three reds =

[ 1 * 4 * 3 * 2] / [ 10 * 9 * 8 * 7]  = 24 / 5040

Probability of one white, one blue, two reds  =

[ 2 * 1 * 4 * 3 ] / [ 10 * 9 * 8 * 7] = 24 / 5040

Probability of one of each color  =

[1 * 2 * 3 * 4 ] /  [10 * 9 * 8 * 7 ]  =  24 / 5040

So...the minimum number is 10

Thanks!

x^2  + x  +  11  =  5x  - 8      rearrange as

x^2 - 4x +  19  = 0

x =   4  ±√  [   (-4)^2  - 4 (1) (19) ]

       _______________________

                    2(1)

x =    4 ±√  [   16  - 76 ]

       _______________

                    2

x =    4 ±√  [   - 60  ]

       ______________

                 2

x =    4 ± 2i√15 

       ______________

                 2

x =  2 ± i√15

angle B, angle K

angle C, angle L

AC, JL

BC, KL

If   S is the scale factor, then  if

l S  l    >   1       we have an expansion       else  if

l S l  <  1 we have a contraction

l -5/4  l   =  5/4   >  1      expansion

l 3/8 l  = 3/8  <  1    contraction

l -1/2 l  = 1/2  <  1    contraction

l 0.5 l  = 0.5 < 1  contraction

The angles of similar triangles are the same, and the ratio of two sides to two corresponding sides are the same, but the length of the sides can be different.

∠K  ≅  ∠T          Notice that  K  and  T  are both in the middle of the name, so this is true.

m∠J  =  m∠S

KL/TU  =  JK/ST

These are the true statements.

Thanks, superman...!!!

But...if the scale factor is truly 25....the image is now 25 * 12 =  300 units long.

the longer side / the shorter side must be the same for rectangles to be similar.

For  A:  100 / 80  =  5/4

For  B:  120 / 100  =  6/5

For  C:  90 / 72  =  5/4

So... only  A  and  C  are similar.

*edit* I changed this answer because I'm pretty sure my old answer was wrong.

The new x  coordinate will be  3 + 1/(4 + 1) * (the positive difference of the x coordinates)    , and

the new  y  coordinate will be  2 + 1/(4 + 1) * (the positive difference of the y coordinates)

the new  x  coordinate   =   3 + 1/5(8 - 3)   =   3 + 5/5   =   4

the new  y  coordinate   =   2 + 1/5(15 - 2)   =   2 + 13/5   =   4.6

So the new coordinates are     ( 4 , 4.6 )

midpoint   =   \((\frac{9+2}{2},\frac{-8+-5}{2})\)   =   \((\frac{11}{2},\frac{-13}{2})\)   =   (5.5, -6.5)

That's Awesome Melody!