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24 / (50/60) =

24 x 60/50    =

1,440/50       =28.80 Km/h - Pam's average cycling speed.

Find a complex number z such that the real part and imaginary part of z are both integers,

and such that zz = 89

I assume, \( z\bar{z} = 89\)

89 has one solution, because 89 is a prime number and \(89 = 1 \pmod{4}\)

\(89 = 5^2 + 8^2\)

so

\(z = 5 + 8i \text{ and } \bar{z} = 5-8i \text{ so } z\bar{z} = 5^2+8^2 = 89\\ \text{or} \\ z= 8 + 5i \text{ and } \bar{z} = 8-5i \text{ so } z\bar{z} = 8^2+5^2 = 89\)

Express 1/(1+1/(1-1/(1-i))) in the form a+bi

where a and b are real numbers.

\(\begin{array}{|rcll|} \hline && \mathbf{\cfrac{1}{1+\cfrac{1}{1-\cfrac{1}{ 1-i }}}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i}{ 1-i }-\cfrac{1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i-1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ \not{1}-i-\not{1} }} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ -i} } \\\\ &=& \cfrac{1}{\cfrac{-i}{-i}+\cfrac{1-i}{-i} } \\\\ &=& \cfrac{1}{\cfrac{-i+1-i}{-i} } \\\\ &=& \dfrac{-i}{-i+1-i} \\\\ &=& \dfrac{-i}{1-2i} \\\\ &=& \left(\dfrac{-i}{1-2i} \right) \cdot \left( \dfrac{1+2i}{1+2i} \right) \\\\ &=& \dfrac{-i(1+2i)}{(1-2i)(1+2i)} \\\\ &=& \dfrac{-i-2i^2}{1-4i^2} \quad & | \quad i^2 = -1 \\\\ &=& \dfrac{-i-2(-1)}{1-4(-1)} \\\\ &=& \dfrac{-i+2}{1+4} \\\\ &=& \dfrac{-i+2}{5} \\\\ & \mathbf{=}& \mathbf{ \dfrac{2}{5}-\dfrac{1}{5}i } \\ \hline \end{array}\)

-2(x+y)+9=1 is this in standard form

\(-2(x+y)+9=1\\ -2x-2y=1-9\\ -2x-2y=-8\\ 2x+2y=8\\ x+y=4\\ \)

dr/dt = 8 cm/s

A = pi*r² so dA/dt = 2pi*r*dr/dt

A = 2pi*25*8 cm²/s → 400pi cm²/s

.

The system of equations \[\frac{3xy}{x + y} = 5, \quad \frac{2xz}{x + z} = 3, \quad \frac{yz}{y + z} = 4\]

has one ordered triple solution $(x,y,z)$.

What is the value of $z$ in this solution? 

\(\begin{array}{|rcll|} \hline \dfrac{3xy}{x + y} = 5, \quad \dfrac{2xz}{x + z} = 3, \quad \dfrac{yz}{y + z} = 4 \\ z=\ ? \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \dfrac{3xy}{x + y} &=& 5 \\\\ & \dfrac{3}{5} &=& \dfrac{x+y}{xy} \\\\ \mathbf{(1)} & \mathbf{\dfrac{3}{5}} &\mathbf{=}& \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \\\\ \hline & \dfrac{2xz}{x + z} &=& 3 \\\\ & \dfrac{2}{3} &=& \dfrac{x+z}{xz} \\\\ \mathbf{(2)} & \mathbf{ \dfrac{2}{3}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \\\\ \hline & \dfrac{yz}{y + z} &=& 4 \\ & \dfrac{1}{4} &=& \dfrac{y+z}{yz} \\\\ \mathbf{(3)} & \mathbf{\dfrac{1}{4}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \\ \hline \end{array}\)

z = ?

\(\begin{array}{|lrcll|} \hline (2)+(3)-(1): & \mathbf{ \dfrac{2}{3}}+\mathbf{\dfrac{1}{4}} - \mathbf{\dfrac{3}{5}} &=& \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) \\\\ & \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} &=& \dfrac{2}{z} \\\\ & \dfrac{2}{z} &=& \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} \\\\ & \dfrac{2}{z} &=& \dfrac{2\cdot 20+15-3\cdot 12}{60} \\\\ & \dfrac{1}{z} &=& \dfrac{40+15-36}{120} \\\\ & \dfrac{1}{z} &=& \dfrac{19}{120} \\\\ & \mathbf{z} &\mathbf{=}& \mathbf{\dfrac{120}{19}} \\ \hline \end{array}\)

x = ?

\(\begin{array}{|lrcll|} \hline (1)-(3)+(2): & \mathbf{\dfrac{3}{5}}-\mathbf{\dfrac{1}{4}}+\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} &=& \dfrac{2}{x} \\\\ & \dfrac{2}{x} &=& \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{ 3\cdot 12-15+2\cdot 20}{60} \\\\ & \dfrac{1}{x} &=& \dfrac{36-15+40}{120} \\\\ & \dfrac{1}{x} &=& \dfrac{61}{120} \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{120}{61}} \\ \hline \end{array}\)

y = ?

\(\begin{array}{|lrcll|} \hline (1)+(3)-(2): & \mathbf{\dfrac{3}{5}}+\mathbf{\dfrac{1}{4}}-\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z} + \dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} &=& \dfrac{2}{y} \\\\ & \dfrac{2}{y} &=& \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{ 3\cdot 12+15-2\cdot 20}{60} \\\\ & \dfrac{1}{y} &=& \dfrac{36+15-40}{120} \\\\ & \dfrac{1}{y} &=& \dfrac{11}{120} \\\\ & \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{120}{11}} \\ \hline \end{array}\)

Find all complex numbers z satisfying the equation (z+1)/(z-1) = i

\(\begin{array}{|lrcll|} \hline & \dfrac{z+1}{z-1} &=& i \quad & | \quad z = a+bi \\\\ & \dfrac{a+bi+1}{a+bi-1} &=& i \\\\ & a+bi+1 &=& i(a+bi-1) \\\\ & a+bi+1 - i(a+bi-1)&=& 0 \\\\ & a+bi+1 - ia-bi^2+i&=& 0 \quad & | \quad i^2 = -1\\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & ( a+b+1) + (b-a+1) i &=& 0 \\ \\ \hline (1) & (a+b+1) &=& 0 \\ (2) & (b-a+1) &=& 0 \\ \hline (1) + (2) & 2b+2 &=& 0 \quad & | \quad : 2 \\ & b+1 &=& 0 \\ & \mathbf{b} &\mathbf{=}& \mathbf{-1} \\ \hline (1) - (2) & 2a &=& 0 \quad & | \quad : 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

\(z = a+bi \\ z = 0 -i \\ \mathbf{z = -i}\)

Simplify (i+1)^{3200}-(i-1)^{3200}

\(\begin{array}{|rclrcl|} \hline && \mathbf{(i+1)^{3200}-(i-1)^{3200}} \\ &=& (i+1)^{2\cdot 1600}-(i-1)^{2\cdot 1600} \\ &=& \left( (i+1)^{2} \right)^{1600}- \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i+1)^{2} &=& i^2+2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1+2i+1 \\ && \quad & \quad &=& 2i \\ &=& (2i)^{1600} - \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i-1)^{2} &=& i^2-2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1-2i+1 \\ && \quad & \quad &=& -2i \\ &=& (2i)^{1600} - (-2i)^{1600} \\ &=& (2i)^{1600} - (-1)^{1600}(2i)^{1600} \quad & \quad (-1)^{1600} = 1 \\ &=& (2i)^{1600} - (2i)^{1600} \\ &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

=0.5

=0.25

=-0.375

=-0.6875

=-0.84375

The slope of the line  y = 43x + 6  is  43 .

The slope of a line perpendicular to that is  -1/43

So we want the equation of a line with a slope of  -1/43

that passes through  (-4, -1)  and a general point,  (x, y)  

\(\text{slope}\,=\,\frac{y_2-y_1}{x_2-x_1} \\~\\ -\frac{1}{43}=\frac{y-(-1)}{x-(-4)} \\~\\ -\frac{1}{43}=\frac{y+1}{x+4} \\~\\ (-\frac{1}{43})(x+4)=y+1 \\~\\ -\frac1{43}x-\frac4{43}=y+1 \\~\\ -\frac1{43}x-\frac4{43}-\frac{43}{43}=y \\~\\ y=-\frac1{43}x-\frac{47}{43}\)

The larger triangle will be  (scale factor)^2  times as large as the smaller triangle

The scale factor is 2...so....the larger triangle will be (2)^2  =  4 times as large

Not in general

P (≥ 3)  =  P(3)  + P (4)  + P (5)  +  ...  + P (N)

But

P ( > 4)  =  P (5) + P (6)  + P (7) + ..... + P (N)

(i + 1)^3200  =

i^3200 + ai^3199 + bi^3198  + ci^3197 + di^3196 +  .....+ di^4 + ci^3 + bi ^2 + ai + 1

(i - 1)^3200  = 

i^3200 - ai^3199 + bi^3198  - ci^3197 + di^3196 -  .....+ di^4 - ci^3 + bi ^2 - ai + 1

So

(i + 1)^3200  - (i - 1)^3200    leaves

2 [ ai^3199 +  ci^3197 + ......+ ci^3 + ai  ]   =

2  [ a (-i + i)  + c (i + - i)  +  e( -i + i) + g ( i + - i)  +  ....... ]  =  

2 [ a * 0  +  c * 0  +  e * 0  + g * 0  +  ......  ]  =

2 [ 0 ]  =

0

Factor  a^2b^4 - 4b^2 + 2a^2b^2 - 8

Factor the following:
a^2 b^4 - 4 b^2 + 2 a^2 b^2 - 8

Factor terms by grouping. a^2 b^4 - 4 b^2 + 2 a^2 b^2 - 8 = (2 a^2 b^2 + a^2 b^4) + (-4 b^2 - 8) = a^2 b^2 (b^2 + 2) - 4 (b^2 + 2):
a^2 b^2 (b^2 + 2) - 4 (b^2 + 2)

Factor b^2 + 2 from a^2 b^2 (b^2 + 2) - 4 (b^2 + 2):
(b^2 + 2) (a^2 b^2 - 4)

a^2 b^2 - 4 = (a b)^2 - 2^2:
(a b)^2 - 2^2 (b^2 + 2)

Factor the difference of two squares. (a b)^2 - 2^2 = (a b - 2) (a b + 2):
(ab - 2) (ab + 2) (b^2 + 2)  OR (1)

Simplify (i+1)^{3200}-(i-1)^{3200}

= 2^1600  -   2^1600 =0  {Per Mathematica 11] !!!!.

Yes! This helped loads! I seem to get it now. 

Thank you for your time!

:) 

This will be the same area as that inside a circle with a radius of 5

So....the area will  be  ≈  pi * 5^2  ≈    25 pi  units^2

Here's the first one

From the 1st point, 8 segments can be drawn to the other points

From the 2nd point, 7 segments can br drawn to the other points [ a segment to this point has already been drawn from the 1st  point ]

So.....continuing the pattern.....the number of total segments is just the sum of the 1st  eight positive integers  =  [ 8 * 9 ]  / 2  = 36

To see this a little more clearly, note that we are really counting all the different pairs we can form by choosing any 2 points from 9 =  C(9,2)  = 36

Simplify the following:
1/(1/(1 - 1/(-i + 1)) + 1)

Multiply numerator and denominator of (-1)/(-i + 1) by 1 + i:
1/(1/((-(i + 1))/((-i + 1) (i + 1)) + 1) + 1)

(-i + 1) (i + 1) = 1×1 + 1 i - i×1 - i×i = 1 + i - i + 1 = 2:
1/(1/(1 - ((i + 1))/2) + 1)

Put each term in 1 - (i + 1)/2 over the common denominator 2: 1 - (i + 1)/2 = 2/2 - (i + 1)/2:
1/(1/(2/2 + (-i - 1)/2) + 1)

Factor -1 from -i - 1:
1/(1/(2/2 + (-(i + 1))/2) + 1)

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

Multiply the numerator of 1/((2 - 1 - i)/2) by the reciprocal of the denominator. 1/((2 - 1 - i)/2) = (1×2)/(2 - 1 - i):
1/(2/(2 - 1 - i) + 1)

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

Multiply numerator and denominator of 2/(-i + 1) by 1 + i:
1/((2 (i + 1))/((-i + 1) (i + 1)) + 1)

(-i + 1) (i + 1) = 1×1 + 1 i - i×1 - i×i = 1 + i - i + 1 = 2:
1/((2 (i + 1))/2 + 1)

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

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

Multiply numerator and denominator of 1/(i + 2) by 2 - i:
(-i + 2)/((i + 2) (-i + 2))

(i + 2) (-i + 2) = 2×2 + 2 (-i) + i×2 + i (-i) = 4 - 2 i + 2 i + 1 = 5:
(-i + 2)/5

We want to find  x .

First let's find  cos a  using the law of cosines:     c²  =  a² + b² - 2ab cos C

48²  =  14² + 50² - 2(14)(50) cos a

2304  =  196 + 2500 - 1400 cos a

-392  =  -1400 cos a

0.28  =  cos a

And using the Pythagorean identity

0.28² + sin² a  =  1

sin² a  =  0.9216

sin a  =  0.96

Now....let's just look at the little triangle.

sin a  =  opposite / hypotenuse

sin a  =  x / 7

0.96 =  x / 7

7 * 0.96  =  x

6.72  =  x

Not an expert on these....but I believe that we have :

8  = number of trials    p = probability of success on any trial = .2    

  probability of failure on one trial  = 1 - .2  = .8

P(3)  = means "probability of  3 successes in 8  trials "  =

  C(8,3) (.2)^3 (1 - .2)^(8-3)  = C(8,3) (.2)^3 (.8)^5  ≈  14.68%

 P (≤3 ) = means probability of  0 successes in 8 trials + probability of  1 success in 8  trials + probability of  2 successes in 8  trials + probability of  3 success in 8  trials  = 

C (8,0) (.2)^0 (.8)^8  +  C (8,1) (.2)^1 (.8)^7 + C (8,2) (.2)^2 (.8)^6 + C (8,3) (.2)^3 (.8)^5  ≈

94.37%

P ( > 4) = 

C (8,5) (.2)^5 (.8)^3  +  C (8,6) (.2)^6 (.8)^2 + C (8,7) (.2)^7 (.8)^1 + C (8,8) (.2)^8 (.8)^0  ≈

1.04%

Based on this....Can you do the rest ???...if not....let me know....

Here's some more info :  http://www.mathwords.com/b/binomial_probability_formula.htm

Note also  :  P(> 4)  = 1 - P( ≤ 4 )

Solve for y:

y^4 = -16

Take 4^th roots of both sides.

Taking 4^th roots gives 2 (-1)^(1/4) times the 4^th roots of unity:

y = -2 (-1)^(1/4)    or y = 2 (-1)^(1/4)      or y = -2 (-1)^(3/4)       or y = 2 (-1)^(3/4)

Sorry if that didnt make any sense, here try this 1/(1+1/(1-1/(1-i)))

Thanks. I’m sorry if this is too much to ask but could you explain why and how you got those answers? 

Thanks, Mr. O   ....!!!!