MaxWong

http://web2.0calc.com/questions/given-that-a-b-c-a-c-d-and-b-c-d-where-d-is-a-positive

Using CPhill's answer:

a = -d

but d is a positive integer, so minimum value of d is 1;

so maximum value of a = -d <-- so a = -1

$3-\dfrac{2x^2+11x+12}{x^2+7x+12}\\ =3-\dfrac{(2x+3)(x+4)}{(x+3)(x+4)}\\ =3-\dfrac{2x+3}{x+3}\;,\;x\neq-4\\ =\dfrac{3x+9}{x+3}-\dfrac{2x+3}{x+3}\;,\;x\neq-4\\ =\dfrac{x+6}{x+3}\;,\;x\neq-4$

The answer you wanted :)

As the triangles are similar, we can relate the lengths by ratios...:

$\dfrac{1\frac{5}{6}}{2\frac{1}{4}} = \dfrac{x}{1\frac{1}{2}}\\ x = 1\dfrac{5}{6} \times 1\dfrac{1}{2} \div 2\dfrac{1}{4} = 1\dfrac{2}{9}$

First one is FALSE.

I have a counterexample: $2^x=0$

That way, there are no solution for x.

Second one I think is true, but I don't know why...

Arithmetic mean of 4.2 , 2.61 , 3.6:

= $\dfrac{4.2+2.61+3.6}{3}$

= 3.47

Geometric mean of 4.2, 2.61, 3.6

= $\sqrt[3]{4.2*2.61*3.6}$

$\approx 3.405$ (corrected to 3 d.p.)

Such a quick way!

The only formula used by me is:

$\displaystyle \int^{1}_0 x^p(1-x^r)^sdx = \dfrac{s}{p+1+rs}B(\dfrac{p+1}{r},s)$

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)

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

This equation is a straight line which is in slope-intercept form.

You can observe easily that slope = -3, y-intercept = -4 

the graphing? do it yourself.

"Calculus".

Isn't that awesome? LOL