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That is correct! Good job ^-^

I'm not a Physics expert, but  I believe we need this equation, RP

VF^2  =VI^2  + 2 * a * d

VF  = 30m/s

VI = 40m/s

d  = the displacement  [ how far it moves]....so we have

a = acceleration  (in m/s^2) = ???

30^2  = 40^2  + 2 * a (1500)

900  =  1600 + 3000a    subtract 1600 from both sides

-700  = 3000a     divide  both sides by  3000

-700 / 3000  = a   = -7/30 m/s^2 ≈ -0.233 m/s^2  [ this makes sense....we are  decelerating !!! ]

You want to put them in order first from least to greatest.

Boys: 70,71,72,73,73,75,76,78

Girls: 66,66,65,68,68,70,72 for the girls the median is 68.

Median means middle.

For the boys it's a little more tricky because there are 2 middle numbers, 73 and 73. First you get the average of those and then divide it by 2. 

73+73=146

146/2=73 so it would be 73 either way. 

To find the difference of the medians do 73-68=5 in. (c)

Oh it's B

b) 25 - 4x  = 15- 3x + 10 - x   simplify

25 - 4x = 25 - 4x

Same expressions on both sides will give us infinite solutions for  x

 c) 4x + 8  = 2x + 7 + 2x  - 20    simplify

4x + 8 =  4x - 13       subtract 4x from both sides

8  = -13     absurd answer  means that no solutions exist.....

Let D  be the distance...and 40 min  = 2/3 hr

And  rate * time  =  distance.....but the distances are the same...so  ...we can make an equation involving the times

Adding 40min  = 2/3 hr to  the time it takes driving 60 mph  = the time it takes driving 50  mph....so...

D / 60   +  2/3   =  D /  50    rearrange as

2/3=  D/50  - D / 60

2/3  = [ 60 - 50 ] D  / 3000

2/3  = 10D / 3000

2/3  = D/ 300    multiply both sides by 300

300 (2/3)  = D  = 200 miles

Usually about a week I think.

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The tangent line at x = 7 passes through the points (7, f(7)) and (10, 7),

in which case

\(\displaystyle f\;'(7) = \frac{7-f(7)}{10-7}.\)

Hi, I like it when people ask for more explanation.   

The very first line is   M=4D

If I divide both sides by 2 then I get.

\(\frac{1}{2}M=2D\)

\(\dfrac{w+z}{1+w z}= \dfrac{(w+z)(1+w z)^*}{(1+w z)(1+w z)^*} = \\ \dfrac{w+z + w(wz)^* + z(wz)^*}{|1+wz|^2}\)

The denominator is real so we just need to show the numerator is also real

\(\text{Noting that }(wz)^* = w^* z^* \text{ we have}\\ w+z+w(wz)^*+z(wz)^* = \\ w + z + w w^* z^* + z z^* w^* = \\ w + z + 1\cdot z^* + 1 \cdot w^* = \\ (w+w^*)+(z+z^*) = 2Re(w) + 2Re(z) = \\ 2Re(w+z) \in \mathbb{R} \)

Prove that if w, z are complex numbers such that \(|w| = |z| = 1\) and \(wz\ne-1\),

then \(\dfrac{w+z}{1+wz}\)  is a real number.

\(\begin{array}{|rcll|} \hline |w| = 1 ~&\Rightarrow& ~|w|^2 = 1 \\ |z| = 1 ~&\Rightarrow& ~|z|^2 = 1 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{w}{1+wz} + \dfrac{z}{1+wz} \\\\ &=& \dfrac{1}{ \dfrac{1}{w} + z} + \dfrac{1} { \dfrac{1}{z} + w} \quad | \quad \dfrac{1}{w} = \dfrac{\bar{w}}{|w|^2}=\bar{w},\ \quad \dfrac{1}{z} = \dfrac{\bar{z}}{|z|^2}=\bar{z} \\\\ &=& \dfrac{1}{ \bar{w} + z} + \dfrac{1} { \bar{z} + w} \\\\ &=& \dfrac{ \bar{z} + w+\bar{w} + z}{(\bar{w} + z)(\bar{z} + w)} \\\\ &=& \dfrac{ ( w+\bar{w}) + (z+\bar{z} )}{(\bar{w} + z)(w+\bar{z})} \quad | \quad w = a+bi,\quad \bar{w} = a-bi, \quad z=c+di, \quad \bar{z}=c-di \\\\ &=& \dfrac{ ( a-bi+a-bi) + (c+di+c-di )}{(a-bi + c+di)(a+bi+c-di)} \\\\ &=& \dfrac{ 2( a+c) }{[~(a+c)+(d-b)i~][~(a+c)-(d-b)i~]} \\\\ &=& \dfrac{ 2( a+c) }{[~(a+c)^2-(d-b)^2i^2~]} \quad | \quad i^2=-1 \\\\ &=& \dfrac{ 2( a+c) }{ (a+c)^2+(d-b)^2 } \\ \hline \end{array}\)

Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 7.

If the second approximation is found to be x2 = −8, 

and the tangent line to f (x) at x = 7 passes through the point (10, 7),

find f (7).

Newton's method:

\(\begin{array}{|rcll|} \hline x_1 - x_2 &=& \dfrac{f(x_1)}{f'(x_1)} \quad & | \quad x_1 = 7 \quad x_2 = -8 \\\\ 7+8 &=& \dfrac{f(7)}{f'(7)} \\\\ 15 &=& \dfrac{f(7)}{f'(7)} \\\\ f(7) &=& 15f'(7) \quad & | \quad f'(7)=\dfrac{7}{10} \\ f(7) &=& 15\cdot \dfrac{7}{10} \\ \mathbf{f(7)} &\mathbf{=}& \mathbf{10.5} \\ \hline \end{array} \)

\(\text{Let's say I can finish the job in }n \text{ hours} \\ \text{thus I work at a rate of }\dfrac{1~job}{n~hr} \\ \text{You finish the job in (}n-2) \text{ hours}\\ \text{and thus work at a rate of }\dfrac{1~job}{(n-2)~hr}\)

\(\text{When working together the rates combine.}\\ \text{I work 1 hour at a rate of }\dfrac{1~job}{n~hr} \\ \text{and then work 3 hours at a rate of } \left(\dfrac{1}{n}+\dfrac{1}{n-2}\right) \dfrac {job}{hr} \text{completing the job}\)

\(1~job = \left(\dfrac {1~job}{n~hr} \cdot 1~hr+\left(\dfrac 1 n+\dfrac{1}{n-2} \right)\dfrac{job}{hr} \cdot 3~hr\right)\)

dropping units for a moment

\(1 = \dfrac 1 n +\dfrac{2n-2}{n(n-2)}\cdot 3\)

\(n^2 - 2n = 7n-8 \\ n^2 -9n+8 = 0 \\ (n-1)(n-8) = 0 \\ n=1,~8\)

\(\text{n can't be 1 as that would make your partner finish the job in (-1) hours}\\ \text{so it must be that }n=8~ hrs\)

What are the next 3 terms of this sequence:

1, 2, 3, 8, 15, 48, 105, 384, 945, 3 840..........etc.

\(\begin{array}{|rcll|} \hline \mathbf{\text{Sequence:} }&& \mathbf{1,\ 2,\ 3,\ 8,\ 15,\ 48,\ 105,\ 384,\ 945,\ 3840, \ldots} \\ \\ a_1 &=& 1 \\ a_2 &=& 2 \\ a_3 &=& 3 \\ a_4 &=& 8 \\ a_5 &=& 15 \\ a_6 &=& 48 \\ a_7 &=& 105 \\ a_8 &=& 384 \\ a_9 &=& 945 \\ a_{10} &=& 3840 \\ \ldots \\ \mathbf{a_{n} }&\mathbf{=}& \mathbf{n!!} \\ \\ \hline \\ a_{11} &=& 11!! \\ \mathbf{a_{11}} &\mathbf{=}& \mathbf{10395} \\\\ a_{12} &=& 12!! \\ \mathbf{a_{12}} &\mathbf{=}& \mathbf{46080} \\\\ a_{13} &=& 13!! \\ \mathbf{a_{13}} &\mathbf{=}& \mathbf{135135} \\ \hline \end{array}\)

Formula \(n!!\):

\(\large{ \begin{equation} n!! = \begin{cases} 2\cdot 4\cdot 6\cdot \ldots \cdot (n-4)\cdot (n-2)\cdot n \text{,} & \text{if } \ n \ \text{ is even} \\ 1\cdot 3\cdot 5\cdot \ldots \cdot (n-4)\cdot (n-2)\cdot n \text{,} & \text{if } \ n \ \text{ is odd} \\ \end{cases} \end{equation} }\)

Example:

\(\begin{array}{|rcll|} \hline 0!! &=& 1 \\ (-1)!! &=& 1 \\ 9!! &=& 1\cdot 3 \cdot 5 \cdot 9 \\ 10!! &=& 2\cdot 4 \cdot 6 \cdot 8 \cdot 10 \\ 19!! &=& 1\cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19 \\ 20!! &=& 2\cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 \cdot 18 \cdot 20 \\ \hline \end{array}\)

wow, thank you

how did you get the 1/2m = 2d?

Replay of invisable posts

Post #1: deleted because it doesn’t make sense.

-------------------------------

Here’s your deleted post:

Let the speed of the plane from A to B = S, then:

Its speed from B to A =S - 100

48/60 =0.8 hours

Distance =Speed x Time

4.8S =D.......................................(1)

5S - 500=D..................................(2)

4.8S =5S - 500

S  = 2,500 mph-which doesn't make sense!!!

Distance =4.8 x 2,500 =12,000 miles.

The reason it doesn’t make sense is because you lack the prerequisites.

Sounds right to me :)

Thank you so much that makes sense!

Thank You!!

Thanks!!

Ok.. if we have a resistance of R ohms in a piece of perfect wire we will hve R/2 ohms in half that length of wire.

If we pass 12 volts through and measure 100 mA of current we have 12/(0.1) = 120 Ohms of resistance in the wire.

If we cut it in half we will only have 60 Ohms of resistance in the wire.

So if we pass 24 volts through it we will measure 24v/60ohs = 0.4 Amps = 400 mA of current

I can also quickly see this by noting that we halve the resistance and double the voltage.

2/(1/2) = 4, so we increase the original current by a factor of 4.

I'D SAY THEY ARE 25...... AND 140!!!!!!!

Let D be the total distance  and T be the total time

So...when on the highway, we have

 (5/6)D  / 60  = (3/4)T     simplify

5D / (6 * 60)  = (3/4)T

5D / 360  = (3/4)T

D / 72  = (3/4)T     multiply both sides by  4/3

(4/3)D / 72  =  T

4D / 216  = T        divide both sides by 4

D / 216  =  (1/4) T     

And when not on the highway....we have (1/6)D left to travel  and (1/4)  of the time left to travel

Let R  be the rate over  this distance...and we have

(1/6)D  / R   = (1/4)T        sub   D/216 for (1/4)T

(1/6)D / R  = D/216      rearrange as

(1/6)D / (D / 216)  = R

(1/6)D * (216 / D)  = R

36 mph  =  R   =  rate while not on the highway