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You wrote this f:

Commutative Property of Multiplication:        ab = ba

So there has to be one thing multiplied by another thing.  Then the 2 are swapped around on the other side of the equal sign.

(Remember that a dot is a multiply sign)

So which one do you think is right?

OR which ones are you pretty sure are wrong ?

I see, so that was the problem, the meter square...Melody, thank you very much..I really do appreciate...

Here is the commutative law.

Which fits this description?  

You can answer that for me

Chris made just a little error think.

4)    Revenue   = -50x^2 + 1500x + 200000

                   R  = -50x^2 + 1500x + 200000

This is a concave down parabola. The maximum will be at the turning point.   (the vertex)

Normally I'd use calculus for this but I'm not sure if you have done any calculus yet.

The axis of symmetry is    x= -b/2a      (from the quadratic equation)

x= -1500/(2*-50) = 1500/100 = 15

So for the most profit there should be 50+15 = 65 passengers.

There is a little error in the question here.

Hi Juriemagic,

A man is making a garden, The garden is 120 square meters. The garden is to be rectangular. It's going to cost him R150 per square meter for the plant fencing which will run on three of the sides, and R250 per square meter for a wire mesh fence, which will run on the remaining side. The wire mesh fence will cover one of the long sides.

The price is not per square metre becasue we are not given the height of the fence. It is just meant to be a price per metre.

You have already worked out that the dimenstions of the garden are   x and   120/x

x is the length so it has to be the long sides.

So on long side costs  R250 per metre so that is   R250x   (Rand is South African money)

Now the other 3 sides add up to (x  + 120/x  + 120/x) = ( x + 240/x )  and they cost R150 per metre.  So that is   R150( ( x + 240/x ) 

\(\text{Total cost }\\=  250x+ 150( ( x + 240/x ) \\=250x+150x+150*240x^{-1} \\= 400x+36000x^{-1}\)

\(z=0 \text{ is an obvious solution}\\ |z|=0\)

\(\text{Assume }z\neq 0\\ z^5 + z^4+2z^3 + z^2 + z = 0\\ z^4 + z^2 = -z(z^4 + 2z^2+1)\\ z(z^2+1) = -(z^2+1)^2\)

\(\text{if }z^2\neq -1\\ z=-(z^2+1)\\ z^2+z+1=0\\ z = \dfrac{-1\pm\sqrt{-3}}{2}=-\dfrac 1 2 \pm i \dfrac{\sqrt{3}}{2}\\ |z| = \sqrt{\dfrac 1 4 + \dfrac 3 4 }=1\)

\(\text{if }z^2=-1\\ z=\pm i\\ |z|=1\)

Thank you, CPhill !

Find all solutions to

\(\large \sqrt[3]{x + 57} - \sqrt[3]{x - 57} = \sqrt[3]{6}\).

\(\begin{array}{|rcll|} \hline \sqrt[3]{x + 57} - \sqrt[3]{x - 57} &=& \sqrt[3]{6} \\ \left( \sqrt[3]{x + 57} - \sqrt[3]{x - 57} \right)^3 &=& 6 \\ (x + 57)-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2}-(x-57) &=& 6 \\ 114-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2} &=& 6 \\ 108-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2} &=& 0 \quad | \quad : 3\\ 36- \sqrt[3]{(x + 57)^2(x - 57)} + \sqrt[3]{(x + 57)(x - 57)^2} &=& 0 \\ 36- \sqrt[3]{(x^2 - 57^2)(x + 57)} + \sqrt[3]{(x^2 - 57^2)(x - 57)} &=& 0 \\ 36- \sqrt[3]{x^2 - 57^2} \underbrace{\left(\sqrt[3]{x + 57} - \sqrt[3]{x - 57}\right)}_{=\sqrt[3]{6}} &=& 0 \\ \sqrt[3]{x^2 - 57^2}\sqrt[3]{6} &=& 36 \\ (x^2 - 57^2)\cdot 6 &=& 36^3 \\ x^2 - 57^2 &=& 7776 \\ x^2 &=& 7776 +3249 \\ x^2 &=& 11025 \\ \mathbf{x} &\mathbf{=}& \mathbf{\pm 105} \\ \hline \end{array}\)

check \(x=105\):

\(\begin{array}{rcll} \sqrt[3]{105 + 57} - \sqrt[3]{105 - 57} &\overset{?}{=}& \sqrt[3]{6} \\ \sqrt[3]{162} - \sqrt[3]{48} &\overset{?}{=}& 1.81712059283 \\ 5.45136177850 - 3.63424118566 &\overset{?}{=}& 1.81712059283 \\ 1.81712059283 & = & 1.81712059283 \quad \checkmark \\ \end{array}\)

check \(x=-105\):

\(\begin{array}{rcll} \sqrt[3]{-105 + 57} - \sqrt[3]{-105 - 57} &\overset{?}{=}& \sqrt[3]{6} \\ \sqrt[3]{-48} - \sqrt[3]{-162} &\overset{?}{=}& 1.81712059283 \\ - 3.63424118566 -(-5.45136177850 ) &\overset{?}{=}& 1.81712059283 \\ - 3.63424118566 + 5.45136177850 &\overset{?}{=}& 1.81712059283 \\ 1.81712059283 & = & 1.81712059283 \quad \checkmark \\ \end{array}\)

CPhill,

the final answer is 1500/100...and you say it's 150?...I know the answer can impossibly be 15, so how?...please explain, sorry for asking again..thank you kindly..

"Amplitude"  does not apply to the tangent, cotangent, secant or cosecant functions

Thanks to EP for pointing this out to me   !!!!

CPhill,

thank you very very much!!..please note I posted this question again under my username, to be able to track in future, I tried to remove this post after I re-posted, but when I came back to try and remove this post, (I don't even know if it can be done)..I saw someone was busy compiling an answer. so I left it. please forgive me if I had broken any rules...

thank you so much for the answer!!

Looks to be 6/5

1) Ticket cost for 60 passengers  =  4000 - (60 - 50)(50)  =   4000 - 10(50)  = R3500

2)  Price of ticket for over 50  =   R [  4000 - 50x  ] 

3) The revenue, R  = number of passengers  * cost of a ticket 

(50 + x)(4000 - 50x)  =   20000 + 4000x - 2500x - 50x^2   = 200000 + 1500x - 50x^2

4)   In the form    ax^2  + bx + c

The number of passengers that will provide the max revenue  is  -b/[2a]

So we have

R  = -50x^2 + 1500x + 200000

So......the number of passengers providing the max revenue is  : 

(-1500) / (2 * -50)  = -1500/ -100 = 15 passengers =  15 over 50  =  65 passengers

EDIT TO CORRECT A SILLY ERROR !!!

Let the first term =  2000

Let the fifth term = 10125

The exponetial form is

A = A₀ b^(t)        where  A₀  is the first term, t is the month [ Jan = 0 ]  and A is the amount in the "t" month

So.....we want to solve this  for b

10125 = 2000b^(4)

10125 = 2000b^4     divide both sides by 2000

10125 /2000 = b^4      take the 4th root of both sides

1.5  = b = growth factor

Jan  = 2000

Feb  = 2000(1.5) = 3500

Mar = 2000(1.5)^2 = 4500

Apr = 2000(1.5)^3 = 6750

May = 2000(1.5)^4  = 10125

We know that it  is exponential because we have a growth factor being raised to a power  and because the growth is not linear

Excellent, heureka!!!!

Find positive integers \((a,b)\) so that \(\sqrt{37 + 20 \sqrt{3}} = a + b \sqrt{3}\).

\(\begin{array}{|rcll|} \hline \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{a + b \sqrt{3}} \\\\ 37 + 20 \sqrt{3} &=& \left(a + b \sqrt{3}\right)^2 \\ 37 + 20 \sqrt{3} &=& a^2+2ab\sqrt{3} +3b^2 \\ 37 + 20 \sqrt{3} &=& \underbrace{a^2+3b^2}_{=37}+\underbrace{2ab}_{=20} \sqrt{3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & 2ab &=& 20 \\ & ab &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{10}{a}} \\ \hline (2) & a^2+3b^2 &=& 37 \\ & a^2+3\dfrac{10^2}{a^2} &=& 37 \\ & a^2+ \dfrac{300}{a^2} &=& 37 \quad |\quad \cdot a^2 \\ & a^4+ 300 &=& 37a^2 \\ & a^4-37a^2+ 300 &=& 0 \\\\ & a^2 &=& \dfrac{37 \pm \sqrt{37^2-4\cdot 300} }{2} \\ & a^2 &=& \dfrac{37 \pm \sqrt{169} }{2} \\ & a^2 &=& \dfrac{37 \pm 13 }{2} \\\\ & a^2 = 25 &\text{or}& a^2 = 12 \\ & a = \pm 5 &\text{or}& a = \pm 2\sqrt{3} \quad |\quad \text{positive integers } (a,b) \\ & a = 5 &\text{or}& a = 2\sqrt{3} \quad |\quad a \text{ without } \sqrt{3} \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b &=& \dfrac{10}{a} \\ & b &=& \dfrac{10}{5} \\ & \mathbf{b} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

\(\begin{array}{rcll} \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{5 + 2 \sqrt{3}} \\ \end{array}\)

Excellent, FL!!!!!

P(sunshine 3 days in a row)  = (.75)^3  ≈ 0.42 ≈  42%

P (rain at least once during the week)  = 

1 - P( of no rain during the week ) =

1 - (.75)^ 7  ≈ .87 ≈  87%

(a)  If independent   

P(winning and playing at home) = P(winning) * P(playing at home)     ????

             0.2                                 =   0.25         *      0.8       ????

             0.2                                 =   0.2

So....they are independent

(b)   

P ( losing and playing away)   =  P (losing ) * P (playing away)   ????

           0.1                               =    0.7           *    0.15      ????

           0.1                              =     0.105

So....not independent