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Compute

\(4 \cos(50^\circ) - \tan(40^\circ)\),

without using a calculator.  
 

How can I do this?

\(\begin{array}{|rcll|} \hline && \mathbf{4 \cos(50^\circ) - \tan(40^\circ)} \quad | \quad \cos(50^\circ)=\cos(90^\circ-40^\circ)=\sin(40^\circ) \\\\ &=& 4\sin(40^\circ) - \tan(40^\circ) \\\\ &=& 4 \sin(40^\circ)*\dfrac{\cos(40^\circ)}{\cos(40^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{2*2 \sin(40^\circ)\cos(40^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad 2 \sin(40^\circ)\cos(40^\circ) = \sin(80^\circ) \\\\ &=& \dfrac{2\sin(80^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad \sin(80^\circ)= \sin(180^\circ-80^\circ)=\sin(100^\circ) \\\\ &=& \dfrac{2\sin(100^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \ \text{or}\ 2=\dfrac{1}{\cos(60^\circ)} \\\\ &=& \dfrac{\sin(100^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ+60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)\cos(60^\circ)+\cos(40^\circ)\sin(60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)\cos(60^\circ)} {\cos(40^\circ)\cos(60^\circ)}+ \dfrac{\cos(40^\circ)\sin(60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)} {\cos(40^\circ)} + \dfrac{\sin(60^\circ)} {\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \tan(40^\circ) + \dfrac{\sin(60^\circ)} {\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(60^\circ)} {\cos(60^\circ)} \quad | \quad \sin(60^\circ)=\dfrac{\sqrt{3}}{2},\ \cos(60^\circ)=\dfrac{1}{2} \\\\ &=& \dfrac{\sqrt{3}}{2} \above 1pt \dfrac{1}{2} \\\\ &=& \mathbf{\sqrt{3}} \\ \hline \end{array}\)

\(\dfrac{3-z}{z+1} \ge 1\\~\\ z\ne-1\\ If \;\;z<-1\;\;then\\ 3-z\le z+1\\ -2z\le-2\\ z\ge1\\ \)

z cannot be less than -1 and also greater than +1 so no solution here

\( If \;\;z>-1\;\;then\\ 3-z\ge z+1\\ -2z\ge-2\\ z\le1\\ -1

My latex is not rendering. Maybe it will is itself later???

Here is an image of it:

LaTex

\dfrac{3-z}{z+1} \ge 1\\~\\ z\ne-1\\ If \;\;z<-1\;\;then\\ 3-z\le z+1\\ -2z\le-2\\ z\ge1\\ 
\text{z cannot be less than -1 and also greater than +1 so no solution here}\\~\\ 
If \;\;z>-1\;\;then\\ 3-z\ge z+1\\ -2z\ge-2\\ 
z\le1\\ 
-1  (-1,1]

Alternative method:

We know that 

0.333... = 1/3

Multiplying both sides by 3:

0.999... = 1

0.999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... = \(9 \displaystyle\sum_{n = 1}^\infty \dfrac{1}{10^n} \)

By sum of infinite geometric series formula, 0.999... = \(9 \cdot \dfrac{\dfrac1{10}}{1-\dfrac{1}{10}} = 9\cdot \dfrac19 = 1\)

That means \(x^2 + bx + c \leqslant 0\) when \(x\in \left[-2, 3\right]\).

And this also means \(x^2 + bx + c < 0\) when \(x\in \left(-2, 3\right)\).

When \(x^2 + bx + c = 0\), \(x = -2\text{ or }x = 3 \)

So \(x^2 + bx + c = (x + 2)(x - 3) = x^2 - x - 6\).

Comparing coefficients, b = -1 and c = -6.

b + c = -1 + (-6) = -7.

Let p and q be primes such that \(p + q = \left(p - q \right)^3\).  Find p and q.

\(\begin{array}{|rcll|} \hline p + q &=& \left(p - q \right)^3 \quad | \quad p = 5,\ q= 3\\ 5 + 3 &=& \left(5 - 3 \right)^3 \\ 8 &=& 2^3 \\ 8 &=& 8\ \checkmark \\ \hline \end{array}\)

Let n and k be positive integers such that n<10^6  and
\(\dbinom{13}{13} + \dbinom{14}{13} + \dbinom{15}{13} + \dots + \dbinom{52}{13} + \dbinom{53}{13} + \dbinom{54}{13} = \dbinom{n}{k}\)

Let n and k be positive integers such that n<10^6  and
\(\dbinom{13}{13} + \dbinom{14}{13} + \dbinom{15}{13} + \dots + \dbinom{52}{13} + \dbinom{53}{13} + \dbinom{54}{13} = \dbinom{n}{k}\)

If x, y, and z are positive real numbers satisfying the system above, then find x, y, and z.
\(x + y + xy = 19, \quad y + z + yz = 29, \quad z + x + zx = 23\)

\(\begin{array}{|rcll|} \hline \mathbf{x + y + xy} &=& \mathbf{19} \\ x (1+y)+ y &=& 19 \\ x (1+y)+ y+1-1 &=& 19 \\ x (1+y)+ (1+y) &=& 20 \\ \mathbf{(1+y)(1+x)} &=& \mathbf{20}\quad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{y + z + yz } &=& \mathbf{29} \\ y (1+z)+ z &=& 29 \\ y (1+z)+ z+1-1 &=& 29 \\ y (1+z)+ (1+z) &=& 30 \\ \mathbf{(1+z)(1+y)} &=& \mathbf{30}\quad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{z + x + zx } &=& \mathbf{23} \\ z (1+x)+ x &=& 23 \\ z (1+x)+ x+1-1 &=& 23 \\ z (1+x)+ (1+x) &=& 24 \\ \mathbf{(1+x)(1+z)} &=& \mathbf{24}\quad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(1)*(3)}{(2)}: & \dfrac{(1+y)(1+x)(1+x)(1+z)} {(1+z)(1+y)} &=& \dfrac{20*24}{30}\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ & (1+x)(1+x)^2 &=& \dfrac{20*24}{30} \\ & (1+x)^2 &=& 16 \\ & 1+x &=& 4 \quad | \quad x > 0 !\\ & \mathbf{x} &=& \mathbf{3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)*(1)}{(3)}: & \dfrac{(1+z)(1+y)(1+y)(1+x) } {(1+x)(1+z) } &=& \dfrac{30*20}{24} \\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ & (1+y)(1+y)^2 &=& \dfrac{30*20}{24} \\ & (1+y)^2 &=& 25 \\ & 1+y &=& 5 \quad | \quad y > 0 !\\ & \mathbf{y} &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(3)*(2)}{(1)}: & \dfrac{(1+x)(1+z)(1+z)(1+y)} {(1+y)(1+x)} &=& \dfrac{24*30}{20} \\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ & (1+z)(1+z)^2 &=& \dfrac{24*30}{20} \\ & (1+z)^2 &=& 36 \\ & 1+z &=& 6 \quad | \quad z > 0 !\\ & \mathbf{z} &=& \mathbf{5} \\ \hline \end{array}\)

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As you became known to others, you would get priority treatment. This would be very beneficial to you.

Plus you could send private messages to make sure the answerer of your choice at least sees your question.

The only reason that I saw this question is that it had been automatically flagged.

I noticed its value when I chose to unflag it.

Now, since you are not a member you may not even realize that your question is no longer flagged and it has been answered.

Membership is free and easy and you will not be bombarded with any unsolicited advertising or anything unpleasant like that.

These questions can always be looked at in a number of different ways.

I got the same answer as you, I think your answers are fine.

Here is my logic (possibly the same as yours)    But definitely more simply displayed.

1) Put down the N

2) 4 places available for the two Os so that is   4C2 = 6 ways

3) 4 distinct letters left for 4 distinct places so that is 4! = 24 ways.

6*24 = 144 possible ways     (where rotations are classed as the same)

How are we to interpret 

as a cross-product, as element by element multiplication or what?

I get 598 as the largest.

Fifteen unit circles are inscribed in an equilateral triangle in such a way that
each circle is externally tangent to its neighbors.
What is the area of the triangle?

\(\mathbf{a=\ ? }\)

\(\begin{array}{|rcll|} \hline A &=&\rho * s \quad | \quad s=\dfrac{a+a+a}{2},\ \rho = 1 \\ A &=&1 *\dfrac{3a}{2} \\ A &=& \dfrac{3a}{2} \\ && \boxed{ 2A = a^2*\sin{60^\circ},\ \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A = a^2*\dfrac{\sqrt{3}}{2} \\ A = a^2*\dfrac{\sqrt{3}}{4} }\\ a^2*\dfrac{\sqrt{3}}{4} &=& \dfrac{3a}{2} \\ a*\dfrac{\sqrt{3}}{4} &=& \dfrac{3}{2} \\ a &=& \dfrac{3}{2} *\dfrac{4}{\sqrt{3}} \\ a &=& \dfrac{6}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ a &=& \dfrac{6\sqrt{3}}{3} \\ \mathbf{a} &=& \mathbf{2\sqrt{3}} \\ \hline \end{array}\)

The area of the triangle:

\(\begin{array}{|rcll|} \hline 2A &=& \left(8+2\sqrt{3} \right)^2 \sin{60^\circ} \quad | \quad \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A &=& \left(8+2\sqrt{3} \right)^2 \dfrac{\sqrt{3}}{2} \\ 4A &=& \sqrt{3}\left(8+2\sqrt{3} \right)^2 \\ 4A &=& \sqrt{3}(64+32\sqrt{3}+12) \\ 4A &=& \sqrt{3}(76+32\sqrt{3}) \\ 4A &=& 96+76\sqrt{3} \quad | \quad : 4 \\ \mathbf{A} &=& \mathbf{24+19\sqrt{3}} \\ \mathbf{A} &=& \mathbf{56.9089653438} \\ \hline \end{array}\)

The area of the triangle is \(\approx \mathbf{56.9}\)

If I cut an equilateral triangle with a perimeter of 6cm into 4 equal size equilateral triangles, what is the total perimeter of these 4 small equilateral triangles.

Hello Guest!

I took the liberty of using cm and corrected 6 to 4. Sorry.

\(4\times 3\times\frac{6cm}{6}=12cm\)

The total perimeter of these 4 small equilateral triangles is 12cm.

  !

angle ADB = 90 degrees because AB is the diameter, and AD and BD are the legs of a triangle ABD

If the perimeter of the sector shown in the figure is 18, with radius 4. What is the angle of the sector in radians?

Hello Guest!

\(P= r\cdot \alpha+2r\\ 18=4\cdot \alpha+8\\ 4\alpha=10\\\color{blue}\alpha=2.5\\ \alpha=2.5\times\frac{\pi}{3.1415926}\)

\(\alpha\approx 0.79577\cdot \pi\)

  !

A = 14.22724134 units squared

You can start by doing 6x6, which is 36. If you multiply it by 6 again, the units digit is still six. So, the last digit is always going to be 6.

6^1 = 6

--------------- the stuff in between the two dash lines is one cycle of the pattern

6^2 = 36

6^3 = 216

6^4 = 1296 (by the way, for these you don't have to multiply the whole thing out, you just need to multiply the last two digits of each result by 6 until you see a pattern lol)

6^5 = 7776

6^6 = 46656

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

6^7 = 279936

And yay! We've arrived back at the start of the pattern, with the tens digit being 3. There's 99 numbers from 2-100, and 99/5 is 19R4 and the 4th tens digit in the pattern cycle is 7.

So the last two digits are 7 and 6.

Thanks for responding but your answers aren't correct :( 

By Pythagoreas, AB = sqrt(13^2 + 19^2)/2 = sqrt(530)/2.

(x,y,z) = (1/2,1/3,1/6)

The smallest possible surface is 6*25 + 4*16 = 214.