asinus

2x^2+3x-20/ x^2+6x+8 {nl} Factorisation

\(2x^2+3x-20 =0\)

a          b          c

\(x = {-3 \pm \sqrt{3^2-4*2*(-20)} \over 2*2}\)

\(x=-\frac{3}{4}\pm\sqrt{169}\)

\({\color{blue}x1=12.25 \ \ \ x2=-13.75}\)

\(x^2+6x+8\)

          p         q

\(x=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}\)

\(x=-3\pm\sqrt{9-8}\)

 \( \( {\color{blue}\frac{2x^2+3x-20}{x^2+6x+8}}\) = \({\color{blue}\frac{(x-12.25)\times(x+13.75)}{(x+2)\times(x+4)}}\)\)

 \( {\color{blue}\frac{2x^2+3x-20}{x^2+6x+8}}\) = \({\color{blue}\frac{(x-12.25)\times(x+13.75)}{(x+2)\times(x+4)}}\)

wot is the power of 147

\({\color{blue}"The \ power \ of \ 147" means \ to \ raise \ a \ number, }\)

\({\color{blue}a \ term, \ a \ variable, \ a \ constant \ into \ the \ 147th \ power.}\)

  !

pie to the power of 7 multiplied by 23

\({\color{blue}{\pi}^7\times 23=69466.7442389}\)   !

what is the are of a rectangle that is square root of 70 by square root of 28. in its simplest form

rectangle: square root of 70 by square root of 28

\(\sqrt{70} \times \sqrt{28} =\sqrt{70\times 28}\)      !

\(A=\sqrt{1960} =44.272\)            !

How do you solve this problem x/3 - (3-x)/6 - (x+2)/8 = (3x-1)/6 ?

\(\frac{x}{3}-\frac{3-x}{6}-\frac{x+2}{8}=\frac{3x-1}{6} \)  

                                           \(Common \ denominator = 24\)

\(\frac{x\times 8-4\times (3-x)-3\times (x+2)}{24} =\frac{4\times (3x-1)}{24} \)

\(8x-12+4x-3x-6=12x-4 \)

\(-3x=14\)

\({\color{blue}x=-\frac{14}{3}}\)

\({\color{blue}x=- 4 \frac{2}{3}}\)

\({\color{blue}x=-4.666\overline{6}}\)      !

Hello Guest!

Where is f(x) = (1/32)sqrt(x) + 1/x, x>0, concave up and concave down?

\({\color{blue}f(x)=\frac{1}{32}\times\sqrt{x}+\frac{1}{x} }\)          x > 0

The graph is downwards convex in the range x> 0.

Greeting asinus :- )   !

2/3/2

If no brackets are set, the order is calculated.

\({\color{blue}2:3:2 = \frac{2}{3}:2 =\frac{2}{6}=\frac{1}{3}}\)    !

Hello Guest!

a solid pyramid of height 40 cm with a square base of sides 30 cm each is put into a cubicle tank of sides 40 cm each. the tank is then filled with water. If the pyramid is removed, find the depth of water in the tank.

V_Tank = \(40^3 cm^3=64000cm^3\)

V_Py = \(\frac{1}{3}\times 30^2cm^2\times 40cm=12000cm^3\)

V_Rest = V_Tank - V_Py = \(64000cm^3-12000cm^3=52000cm^3\)

h = V_Rest / A_Tank = \( \frac{52000cm^3}{40^2cm^2}{\color{blue}=32,5cm}\)

\({\color{blue}After \ removing \ the \ pyramid \ from \ the \ tank, }\)

\({\color{blue}the \ water \ is \ 32.5cm \ high \ in \ the \ tank.}\)

Greeting asinus :- )   !

if A= pi r squared 2 find A when r=34 cm

\({\color{blue}A=r^2\times \pi=34^2cm^2\times\pi=3631.681cm^2}\) !

What is 666 to the power of 666

\({\color{blue}666^{666}=e^{4329.8589205798991406}}\) ≈ ∞   \(!\)