All Answers 358087 Answers

if you mean i/(i+1) , the answer is;

\(\sqrt{-1}/(\sqrt{-1}+1)\)

now, multiply each side with \((\sqrt{-1}-1)\)

you will get \((-1-\sqrt{-1})/(-1-1)\)

that is \((-1-\sqrt{-1})/-2\)

get rid of the ((-)) ; the answer is most basic this;

\((1+\sqrt{-1})/2\)

\((1+i)/2\)

if you mean (i/i)+1, that is 1+1=2 -_-

Thanks ElectricPavlov !

\(\frac{i}{i+1}=\frac {i(i-1)}{(i+1)(i-1)}=\frac {-1-i}{-1-1}=\frac{1+i}{2}=\frac12+\frac 12i\)

as the same as \(a+bi\)

i/(i+1)=1/2+i/2

y=p-2qx^2 

y-p = -2qx^2

-(y-p) / 2q = x^2

x= sqrt(( p-y)/2q)

\(x^2-10x=0\\ x(x-10)=0\\ x_1=0\\ x_2=10\)

Either complete the square or use the Quadratic formula to find x = 0 or 10

Solveit's answer (below) is simpler....good job solveit!

1+1=2

Can someone explain how -1+1/2x5 on top of 1+1/2 is 3/2?

You should bracket them, so that we know what goes first, like this:[-1+ 1/(2x5)] / [1+ 1/2]

As it stands, it doesn't add up to 3/2. More like: -3/5=-.6

\(553\div5=110.6\)

 there is 110 numbers wich is devisible by 5

\(553\div3=184.333\)

there is 184 numbers wich is devisible by 3

\(553\div15=36.9\)

there is 36 numbers wich is devisible by both 5 and 3

now let s find sum of devisors of 5:

\(5+5\times2+5\times3+5\times4\dots+5\times110\\ 5(1+2+3+4\dots+110)\\ 5(55.5\times110)=30\ 525\)

sum of devisors of 3:

\(3(92.5\times184)=51\ 060\)

sum of devisors of both 5 and 3

\(15(18.5\times36)=9990\)

\(30\ 525+51\ 060-9990=71\ 595\)

553 -1 =552

552/3 =184 terms

Sum of 184 "3's" =51,060.

553 - 3=550

550/5 =110 terms

Sum of 110 "5's" =30,525

So it _was_ too easy.

37 ?

\(-1+\frac1{2\times5}=-1+\frac{1}{10}=\frac{-10+1}{10}=-\frac{9}{10}\)

\(1+\frac12=\frac{2+1}{2}=\frac32\)

\(\begin{array}{rcll}2^3 \times 2^x &=& 2^{3x}\\ 2^{3+x}&=&2^{3x}\\ 3+x&=&3x\\ 3&=&2x\\ 1.5 &=&x \end{array}\)

undefined

1- ∫x dx = x^2/2+C
2- ∫x^2 dx =  x^3/3+C
3- ∫x^3 dx =x^4/4+C
 

1.   1/2 x^2

2.   1/3 x^3

3    1/4 x^4

Agree with guest   There MAY be a constant added to any of the above answers

Case 1: 6 women. Then (6,0) is the only possible ordered pair.

Case 2: 5 women and 1 man. Then (6,2) is the only possible ordered pair: all 6 people must be sitting next to a woman, and 2 of the women are sitting next to the man.

Case 3: 4 women and 2 men. Then, if the men sit next to each other or opposite each other, our ordered pair is (6,4); if they sit one apart, our ordered pair is (5,3).

Case 4: 3 women and 3 men. Then if the arrangement is WWWMMM, WWMWMM, or MMWMWW, our ordered pair is (5,5). If it is WMWMWM, our ordered pair is (3,3).

Case 5: 2 women and 4 men. By symmetry with Case 3, the possible pairs are (4,6) and (3,5).

Case 6: 1 woman and 5 men. By symmetry with Case 2, the only possible pair is (2,6).

Case 7: 6 men. (0,6)

This yields a total of 10 ordered pairs.

2^3 * 2^x = 2^(3x)

8*2^x = 2^(3x)           Take log of both sides to get

log8 + xlog2 = 3x log2  divide by log2

log8/log2 + x = 3x

log8/log2 = 2x

1/2 log8/log2 = x = 1.5

Hallo Guest!

1/3.438^14 = 3.102542841088 x10^-8

asinus :- )

  !

1-65/80 = 0.1875

Answer: decresed by 18.75 %