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Solve for x
\(3\cos(x) +2\sin^2 (x)=0\)

\(\begin{array}{|rcll|} \hline 3\cos(x) +2\sin^2(x) &=& 0 \\ 3\cos(x) &=& -2\sin^2(x) \quad &| \quad : (-2) \\ -1.5\cos(x) &=& \sin^2(x) \\ \hline \end{array}\)

Substitute:

\(\begin{array}{|rclcrcl|} \hline t=\tan\left(\dfrac{x}{2}\right) \\ \hline \cos(x) &=& \dfrac{1-t^2}{1+t^2} && \sin(x) &=& \dfrac{2t}{1+t^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline -1.5\cos(x) &=& \sin^2(x) \quad | \quad \cos(x) = \dfrac{1-t^2}{1+t^2},\ \sin(x) = \dfrac{2t}{1+t^2} \\ -1.5\dfrac{(1-t^2)}{(1+t^2)} &=& \dfrac{4t^2}{(1+t^2)^2} \\\\ -1.5 (1-t^2) &=& \dfrac{4t^2}{ 1+t^2 } \\\\ -1.5 (1-t^2)(1+t^2) &=& 4t^2 \\ -1.5 (1-t^4) &=& 4t^2 \\ -1.5+1.5t^4 &=& 4t^2 \\ \mathbf{ 1.5t^4 -4t^2-1.5} &=& \mathbf{0} \\\\ t^2 &=& \dfrac{4\pm \sqrt{16-4*(1.5)*(-1.5)} }{2*1.5} \\ t^2 &=& \dfrac{4\pm \sqrt{16+9} }{3} \\ t^2 &=& \dfrac{4\pm \sqrt{25} }{3} \\ t^2 &=& \dfrac{4\pm 5 }{3} \\\\ t^2_1 &=& \dfrac{4+5}{3} \\ t^2_1 &=& \dfrac{9}{3} \\ \mathbf{ t^2_1 } &=& \mathbf{3} \\\\ t^2_2 &=& \dfrac{4-5}{3} \\ \mathbf{ t^2_2 } &=& \mathbf{ -\dfrac{1}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcl|rcl|} \hline \mathbf{x=\ ?} && &\mathbf{x=\ ?} \\ \hline \tan\left(\dfrac{x}{2}\right) &=& t_1 & \tan\left(\dfrac{x}{2}\right) &=& t_2 \\ \tan\left(\dfrac{x}{2}\right) &=& \sqrt{3} & \tan\left(\dfrac{x}{2}\right) &=& \sqrt{-\dfrac{1}{3}} \qquad (\text{the root is imaginary !}) \\ \hline \end{array} \)

\(\begin{array}{|rcl|rcl|} \hline \tan\left(\dfrac{x}{2}\right) = \sqrt{3} && & \tan\left(\dfrac{x}{2}\right) = \sqrt{3} \\ \hline \tan\left(\dfrac{x}{2}\right) &=& +\sqrt{3} & \tan\left(\dfrac{x}{2}\right) &=& -\sqrt{3} \\ \dfrac{x}{2} &=& \arctan(\sqrt{3})+180^\circ k \quad k\in\mathbb{Z} & \dfrac{x}{2} &=& \arctan(-\sqrt{3})+180^\circ k \quad k\in\mathbb{Z} \\ \dfrac{x}{2} &=& 60^\circ+180^\circ k \quad k\in\mathbb{Z} & \dfrac{x}{2} &=& -60^\circ+180^\circ k \quad k\in\mathbb{Z} \\ \mathbf{x} &=& \mathbf{120^\circ+360^\circ k \quad k\in\mathbb{Z}} & x &=& -120^\circ+360^\circ k \quad k\in\mathbb{Z} \\ & & & x &=& 360^\circ-120^\circ+360^\circ k \quad k\in\mathbb{Z} \\ & & & \mathbf{x} &=& \mathbf{240^\circ+360^\circ k \quad k\in\mathbb{Z}} \\ \hline \end{array}\)

A rancher with 240 meters of fence intends to enclose a rectangular region along a river​ (which serves as a natural boundary requiring no​ fence). Find the maximum area that can be enclosed.

That would probably be my best guess

biggest = 27*28

smallest= 15*40

Consider a regular 23-gon. How many diagonals does the regular polygon have?

\(\begin{array}{|rcll|} \hline \text{diagonals} &=& (22+ 21+20+\ldots+3+2+1) - 23 \\ &=& \dfrac{1+22}{2}*22 - 23 \\ &=& 23*11 - 23 \\ &=& 23*10 \\ \mathbf{\text{diagonals}} &=& \mathbf{230} \\ \hline \end{array} \)

Angle of incidence is the angle between the normal to the mirror and the incident ray, so here that is 90 - 30 or 60 degrees.

The angle of reflection is the same, namely, 60 degrees. 

Total angle turned through by the ray is 360 - 60 - 60 = 240 degrees.

Write \(tan^2(\phi)\) in terms of \(cos(\phi)\).

\(\begin{array}{|rcll|} \hline \sin^2(\phi)+\cos^2(\phi) &=& 1 \quad & | \quad : \cos^2(\phi) \\\\ \dfrac{\sin^2(\phi)}{\cos^2(\phi)}+ \dfrac{\cos^2(\phi)}{\cos^2(\phi)} &=& \dfrac{1}{\cos^2(\phi)} \\\\ \tan^2(\phi) + 1 &=& \dfrac{1}{\cos^2(\phi)} \\ \mathbf{ \tan^2(\phi) } &=& \mathbf{\dfrac{1}{\cos^2(\phi)} -1} \\ \hline \end{array} \)

Kevin is trying to find a white sock in his drawer.

He has 16 white socks, 4 brown socks, and 6 black socks.

What is the probability that he pull out either a black or brown sock, puts it back,

and then pulls out a white sock?

My attempt:

A freight train from city A to city B and a passenger train from city B to city A left the cities at the same time, at 10:00 AM, heading towards each other. The distance between the cities is 360 miles. The freight train is travelling at 50 mph, the passenger train is travelling at 70 mph.
At what time will the trains pass each other?

Ein Güterzug von Stadt A nach Stadt B und ein Personenzug von Stadt B nach Stadt A verließen die Städte gleichzeitig um 10:00 Uhr und fuhren aufeinander zu. Die Entfernung zwischen den Städten beträgt 360 Meilen. Der Güterzug fährt mit 50 mph, der Personenzug mit 70 mph.
Wann fahren die Züge aneinander vorbei?

\(v_{ft}\cdot t+v_{pt}\cdot t=s\\ t(v_{ft}+v_{pt})=s\\ t=s/(v_{ft}+v_{pt})\\ t=360ml/((50+70)mph)\)

\(t=3h\)

\(10\ AM+3h=1\ PM\)

At 1:00 PM the trains pass each other.

  !

The endpoints of a diameter of a circle are (6,2) and (−4,7) .

What is the standard form of the equation of this circle?

Center of the circle:

\(\begin{array}{|rcll|} \hline C &=& \dfrac{\binom{6}{2}+\binom{-4}{7}} {2} \\ C &=& \dfrac{\binom{6-4}{2+7}} {2} \\ C &=& \dfrac{\binom{2}{9}} {2} \\ C &=& \binom{1}{4.5} \\ \hline \end{array} \)

The center of the circle is \((1,~4.5)\)

The radius of the circle:

\(\begin{array}{|rcll|} \hline r^2 &=& (6-1)^2+(2-4.5)^2 \\ r^2 &=& 5^2+(-2.5)^2 \\ r^2 &=& 25 + 6.25 \\ \mathbf{r^2} &=& \mathbf{31.25} \\ \hline \end{array}\)

The equation of this circle: \(\mathbf{(x-C_x)^2 +(y-C_y)^2=r^2 }\)

\(\begin{array}{|rcll|} \hline \mathbf{(x-1)^2 +(y-4.5)^2=31.25 } \\ \hline \end{array} \)

Calculate

\(\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot3\cdot4}+\ldots+\dfrac{1}{99\cdot100\cdot101}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot3\cdot4}+\ldots+\dfrac{1}{99\cdot100\cdot101} } \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n(n+1)(n+2)} \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n}\left( \dfrac{1}{(n+1)(n+2)} \right) \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n}\left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{n(n+1)}-\dfrac{1}{n(n+2)} \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{n}-\dfrac{1}{n+1} - \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2} \right) \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{2n}-\dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \right) \\ &=& \mathbf{\sum \limits_{n=1}^{99} \dfrac{1}{2n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\sum \limits_{n=1}^{99} \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^{99} \dfrac{1}{2n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\sum \limits_{n=1}^{99} \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{2} \sum \limits_{n=1}^{99} \dfrac{1}{n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\dfrac{1}{2}\sum \limits_{n=1}^{99} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{2} \sum \limits_{n=-1}^{97} \dfrac{1}{n+2} -\sum \limits_{n=0}^{98} \dfrac{1}{n+2} +\left( \dfrac{1}{2}* \dfrac{1}{100} + \dfrac{1}{2}* \dfrac{1}{101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}* \dfrac{1}{1} +\dfrac{1}{2}*\dfrac{1}{2} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && -\sum \limits_{n=0}^{98} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}+ \dfrac{1}{4} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && - \dfrac{1}{2} - \sum \limits_{n=1}^{98} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}+ \dfrac{1}{4} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && - \dfrac{1}{2} - \dfrac{1}{100} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{2}+ \dfrac{1}{4} - \dfrac{1}{2} - \dfrac{1}{100}+\dfrac{1}{200} + \dfrac{1}{202} \\ && + \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} + \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{100}+ \dfrac{1}{200} + \dfrac{1}{202} + \sum \limits_{n=1}^{97} \dfrac{1}{n+2} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{100}+ \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{4} - \dfrac{2}{200}+ \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{2} \left( \dfrac{1}{2} - \dfrac{1}{100} + \dfrac{1}{101} \right) \\\\ &=& \dfrac{1}{2} \left( \dfrac{98}{200} + \dfrac{1}{101} \right) \\\\ &=& \dfrac{1}{2} \left( \dfrac{98*101+200}{200*101} \right) \\\\ &=& \dfrac{49*101+100}{200*101} \\\\ &=& \mathbf{ \dfrac{5049}{20200} } \\ \hline \end{array}\)

Thanks for the help. Just want to double-check with the master of all math (), the Maximum Value of P is 756 and the minimum value of P is 600 right?

Maximum= (10+7+5+4+1)(9+8+6+3+2)=756

Minimum= (10+9+8+7+6)(5+4+3+2+1)=600

This is correct right?

Thanks,

PharaoCarl

One train gets a 60 mile head start...its distance from Chicago will be given by

      60  +  60 x     miles      where x = hours after 2nd train starts

2nd train distance = 80 x     when are these two distances equal?

60 + 60x = 80x

x = 3 hours after sencond train starts      4 hours after first train started

Another way:

train 1 has 60 mile head start      train 2 travels 20 m/hr faster     to close that 60 mile head start will require  60 m / 20m/hr = 3 hours

apples =   (50+x) (800-10x)    find max

                40 000 - 500x + 800x - 10x^2

                 -10 x^2  + 300 x + 40000                    max occurs at x = -b/2a =   - 300/-20 = 15 additional trees

50 trees + 15 trees = 65 trees      output =  -10(15^2) + 300 (15)+(40000) = 42250 apples from 65 trees

I already explained it. There is very little to explain.

Have you come up with an answer based on what I already told you.

Or do you just want to copy the exact answer of someone else.

Answer on the original thread. It is rude to repost just because the response does not meet with your approval.

Could you please explain it to me? I've had trouble answering these types of questions and an explanation would very much appreciate it!

Thanks

oops... I tried to search it up but came up with nothing. 

It means 

the measurement of angle bca.

I have only seen it a few times but that was what it meant.