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(2/(20 + 1) ) = ?

\(\frac{2}{20+ 1} = \frac{2}{21}= 0.\overline{095238}...\)

E=vh, where v=frequency, h=Planck's constant

Convert 8.9 eV into Joules =1.43×10^-18 J (joules)

1.43 x 10^-18=v x 6.63 x 10^-34

v= 1.43×10^-18 J / [6.63 x 10^-34]

v=2.157 x 10^15 hertz-the frequency of the photon.

E=2.157 x 10^14 x 6.63 x 10^-34

E=1.43 x 10^-18 Joules

E=8.9eV - the energy of the freed electron!. Or "C"

394

Don't listen to the other anwsers.

No, there is actually Three Point Form in Inverse Proportionality.

x*45/67(859-09e3)!0!/2

80992

You're welcome

Don't you think that you should give a little more information instead of just "Dollars obligated"??!!!.

I could interpret that in half a dozen different ways!!. Be more specific as to what you want done.

bc=15

Formula is: A=h*(a+b)/2 where a and b are the two parallel sides.

b=7

a=12

57=h(12+7)/2

h=57*2/19

h=6

Thanks!

There are 3 complex numbers a+bi, c+di,e+fi. If b=3, e=-a-c, and the sum of the numbers is 2i, find d+f.

[a + bi ] + [ c + di] + [ e + fi]  =   2i

Therefore  

a + c + e  = 0  →   a + c + [-a -c]  = 0    which is true

And

b + d + f   = 2    and if b = 3, then

3 + (d + f)   = 2       subtract 3 from both sides

(d + f)   = - 1

Hello Guest!

One light year equals 5.9*10^12 miles.

How many light-years are in 6.79*10^16 miles?

\(\frac{1\ leat\ year}{5.9\times 10^{12}\ miles} = \frac{x}{6.79\times 10^{16}\ miles} \)

 \(x=\frac{6.79\times 10^{16}}{5.9\times 10^{12}}\ leat\ year\)

\(x = 11508.4745763\ leat\ year\)

Greeting asinus :- )

  !

The interior angles of a square meansure 360°, but each one is 90°.

9 base 10 in base 5 is 14. Is that what your asking?

If your asking 9 in base 5, then there's no such thing.

Let the width = W

And the length = W + 4

And l x w  = area     ....so.......

(w + 4) (W)   = 60     simpliify

W^2 + 4W - 60   = 0      factor

(W + 10) (W - 6)  =  0      and setting each factor to 0, W = -10 cm [reject],    or W  = 6cm

And the length is 4 more than this =  10cm

200(5/6)=1000(1/6)

(6.9*10¹⁶)/(5.9*10¹²)=~11,508

f(x) = 17/x

Write this as 

y = 17/x      multiply both sides by x and divide both sides by y

x = 17/y     "exchange"  x and y

y = 17/x      for y, write f^-1(x)

f^-1(x)  = 17/ x         thus, amazingly..........this function is also its own inverse !!!

f(x)=17/x

x=17/f(x)

1/f(x)=x/17

f^-1(x)=f(x)

\(\begin{array}{rll}&&2\\+&2&3\\&2&5&\end{array}\)

Yay I finally learnt how to use \begin {array} and \end {array}!! :D I learnt it all by myself :D

\(\dfrac{rs}{t}+\dfrac{st}{r}+\dfrac{tr}{s}\\ =\dfrac{r^2s^2}{rst}+\dfrac{s^2t^2}{rst}+\dfrac{t^2r^2}{rst}\\ =\dfrac{r^2s^2+s^2t^2+t^2r^2}{rst}\)

\(\text{Let x be }(rs + st + tr)^2\\ x=(rs + st + tr)^2 = r^2s^2+s^2t^2+t^2r^2+2rs^2t+2r^2st+2rst^2\\ \therefore r^2s^2+s^2t^2+t^2r^2=(rs + st + tr)^2 - 2rst(s+r+t)\)

\(\text{Substitute the expression.} \\\dfrac{r^2s^2+s^2t^2+t^2r^2}{rst}\\=\dfrac{(rs+st+tr)^2-2rst(r+s+t)}{rst}\)

\(\text{Substitute the values.}\\ r+s+t = \dfrac{-(-5)}{1}=5\\ rst = \dfrac{-(-9)}{1}=9\\ rs+st+tr = \dfrac{6}{1}= 6\)

\(\dfrac{(rs+st+tr)^2-2rst(r+s+t)}{rst}\\=\dfrac{6^2-2(9)(5)}{9}\\ =4-10\\ =-6\)

\(\therefore \dfrac{rs}{t}+\dfrac{st}{r}+\dfrac{tr}{s}= -6\)

Let t be a root of f(x)=x^3-x+2. Evaluate t^6-t^2+4t.

\( f(x)=x^3-x+2. \qquad Evaluate\quad t^6-t^2+4t.\\ \mbox{since t is a root I can use factor theorem and say ..}\\ t^3-t+2=0\\ t^3=t-2\\ (t^3)^2=(t-2)^2\\ t^6=t^2-4t+4\\ so\\~\\ t^6-t^2+4t\\=t^2-4t+4-t^2+4t\\=4 \)

I have not seen a question quite like this one before :)

The (sum of) interior angles of a square measure 360°

' The coefficients are 1, a and b ... '

Why isn't c a coefficient ?

x tends to 0 ((3+2x)^5-243)/(3x)

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \left( \frac{ (3+2x)^5-243 } {3x} \right) } \\\\ &=& \lim \limits_{x\to 0} { \left( \frac{ \binom{5}{0}3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -243 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ 3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -3^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( ~ \binom{5}{1}3^3\cdot 2+ \binom{5}{2}3^2\cdot 4x+ \binom{5}{3}3^1\cdot 8x^2+ \binom{5}{4}16x^3+ \binom{5}{5}\frac{32}{3}x^4 ~ \right) }\\\\ &=& \binom{5}{1}3^3\cdot 2 \\\\ &=& 5\cdot 27 \cdot 2 \\\\ &=& 270 \\\\ \hline \end{array}\)

See reply #3 to your original post.