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To convert  rads to degrees :

rads  *  180  / pi      ....so we have...

pi / 3  *  180  / pi  =

180 /3  =   60°

And

tan (60°)  = √3  =  tan (pi/ 3 rads )

Thank you, CPhill!

Nice, tertre  !!!

Just the answer: \(\sqrt{3}\)

  By scanning the numerator and the denominator, we can see common numbers. So, 3 gets cancels out, 4, 5, 6, up to b, an unknown number. After all these cancelations, we are left with \(\frac{a}{2}=9\). This means that a, is \(9*2=18\). To find b, we look at our previous fractions, noting that the denominator is one less than the numerator, so that leaves us with: \(b=17\).

Thus, the answer is \(18+17=\boxed{35}\).

In point slope form, the equation is:

y - 2  =  7(x - 1)      Now get this equation into standard form  (   Ax + By  =  C   ).

                            Distribute  7  to the terms in parenthesees.

y - 2  =  7x - 7

                            Add  7  to both sides of the equation.

y + 5  =  7x

                            Subtract  y  from both sides.

5  =  7x - y

7x - y  =  5

Parallel lines  have the same slope

So...we have

y  = m  (x - x1)  + y1      where m = the slope  = -3/4   and  (x1, y1)  = ( 4,2)

y  = (-3/4) ( x - 4)  + 2

y  = (-3/4)x + 3 + 2

y = (-3/4)x + 5      [in slope-intercept  form ]

Perpendicular lines have negative reciprocal slopes.....so....the slope of the line we want will be -7/2

So  we have

y  = m ( x - x1)  + y1          where  m is the slope and  (x1, y1)  = ( 2,4)

y  = (-7/2) ( x - 2)  + 4         simplify

y = (-7/2)x + 7 + 4

y = (-7/2)x + 11    [  in slope-intercept form ]

We have an irregular trapezoid

If we drew a perpendicular line from the top right vertex of the figure to the base......the right half of   the two figures formed would be a right triangle 

The base length of this triangle  is (24 - 10)  = 14  units

And since one of the angles of the right triangle is 45.....the other angle at the top of the triangle is also 45....and in a 45-45-90  right triangle.....the sides opposite the 45° angles are equal....so....the height of this triangle  = the base length  = 14  =  the height of the trapezoid

And the area of a trapezoid  is 

(1/2) height  *  ( sum of the bases)  =

(1/2) (14) (24 + 10)  =

7 * 34  =

238

Here is a picture to help you understand

If we drew a line connecting the two vertices that are furthest from each other, we would create two triangles, and by SSS, these triangles would be congruent.

And  n°  would be the same as the measure of its corresponding angle, which is  142° .   So...

n  =  142

In every quadrilateral, the sum of the interior angles  =  360° ,   so....

m  =  360 - n - 142 - 20  =  360 - 142 - 142 - 20  =  56

m  =  56

Assuming that this is a regular octagon.......

We can break the octagon into 8 congruent triangles like this:

The sum of the 8 central angles  =  360° , and each central angle has the same measure,

so the measure of one central angle  =  360° / 8  =  45°

Draw a height of one of the triangles perpendicular to a side of the octagon. I labeled the height "a".

By the HL theorem, the two smaller triangles that are formed are congruent.

So the measure of the angle formed by the height and the hypotenuse  =  45° / 2  =  22.5°

tan 22.5°  =  1 / a  cm

a tan 22.5°  =  1  cm

a  =  1 / tan 22.5°  cm

area of one of the eight triangles  =  (1/2)(base)(height)

area of one of the eight triangles  =  (1/2)(2)(1/tan 22.5°)   sq cm

area of one of the eight triangles  =  1 / tan 22.5°   sq cm

area of octagon  =  8 * (area of one of the eight triangles)

area of octagon  =  8 * (1 / tan 22.5° )   sq cm

area of octagon  =  8 / tan 22.5°   sq cm

area of octagon  ≈  19.3   sq cm

We have an octagon  with an  apothem  of 8

We have a right triangle formed

The central angle formed by the triangle is 22.5°

So...the base of this triangle  (1/2 side length of the octagon)  is given by 8 tan (22.5°)

The apothem and the base therefore form two legs of the right triangle

And the area of a right triangle  = (1/2) product of the leg lengths

So....the area  of the right triangle is  (1/2) (8) (8 tan 22.5°)

But  we  have  16 of these  congruent right triangles that comprise the area of the octagon

So...the total  area is   (16) (1/2) (8) * 8 tan (22.5°)  =

8^3 * tan (22.5°)  ≈  212.1  units^2

y  = 0.32x + 20.51

If x  = 175, we have

y  = 0.32 (175) + 20.51  

y  = 76.51   ≈  77 hits

Here's a hint: try using induction

(Use induction to prove that if it is possible to erase the numbers 1, 2, 3, ... , n and remain with 0 then it is also possible to erase the numbers 1, 2, ..., n, n+1, n+2, n+3, n+4 and remain with 0)

The shaded figure shown in the image is a parallelogram. 

The area of a parallelogram is base x height. 

In this image, the base is 12 and the height is 6. 

6 * 12 = 72, this is the area. 

I hope this helped,

Gavin.

2.

How many zeroes do we write when we write all the integers from 1 to243  in base 3?

\(\text{Let base $b = 3$} \\ \text{Let digit numbers $= n$ }\)

1. The one digit numbers \((n=1)\) don't have any zeroes. 

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 0\times \dfrac{2\cdot 3^0}{3} = 0 \text{ zeroes}\)

2. The two digit numbers \((n=2)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 1\times \dfrac{2\cdot 3}{3} = 2 \text{ zeroes}\)

3. The three digit numbers \((n=3) \) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 2\times \dfrac{2\cdot 3^2}{3} = 12 \text{ zeroes}\)

4. The four digit numbers \((n=4)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 3\times \dfrac{2\cdot 3^3}{3} = 54 \text{ zeroes}\)

5. The five digit numbers \((n=5)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 4\times \dfrac{2\cdot 3^4}{3} = 216 \text{ zeroes}\)

6. The six digit numbers \((n=6) \) have:

\(243 = 100000_3 = 5 \text{ zeroes}\)

\(\text{Ths sum $= 0 + 2 + 12 + 54 + 216 + 5 = 289$ zeroes } \)

We write 289 zeroes, when we write all the integers from 1 to 243  in base 3

I oppose questioners having answerers "private message" them the answers. This is because of three reasons:

1. 

How are we supposed to know whether or not someone has already answered the question, or are you just expecting 20 people to "private message" you the answer. It's a waste of time for all the effort in composing the answer to satisfy your uncanny request. 

2. 

"Private messaging" you do answer prevents other users from viewing the solution. This way, other people cannot learn, just to satisfy your uncanny request.

3. 

To my knowledge, you do not receive points from sending private messages. Not only you are preventing others from learning, and causing confusion for the answerers, you are completely screwing up the point system. 

SPQ is a straight line. We have sin RPQ = 7/25. What is sin RPS ?

\(\begin{array}{|rcll|} \hline RPS &=& 180^{\circ} - RPQ \\ \sin( RPS ) &=& \sin ( 180^{\circ} - RPQ ) \\ \sin( RPS ) &=& \sin ( RPQ ) \\ \sin( RPS ) &=& \sin ( RPQ ) \quad & | \quad \sin ( RPQ ) = \dfrac{7}{25} \\ \mathbf{\sin( RPS )} & \mathbf{=} & \mathbf{ \dfrac{7}{25} } \\ \hline \end{array}\)

Here are all numbers from 1 to 243 in base 3. Do you think you can count all the zeros??

{1_3, 2_3, 10_3, 11_3, 12_3, 20_3, 21_3, 22_3, 100_3, 101_3, 102_3, 110_3, 111_3, 112_3, 120_3, 121_3, 122_3, 200_3, 201_3, 202_3, 210_3, 211_3, 212_3, 220_3, 221_3, 222_3, 1000_3, 1001_3, 1002_3, 1010_3, 1011_3, 1012_3, 1020_3, 1021_3, 1022_3, 1100_3, 1101_3, 1102_3, 1110_3, 1111_3, 1112_3, 1120_3, 1121_3, 1122_3, 1200_3, 1201_3, 1202_3, 1210_3, 1211_3, 1212_3, 1220_3, 1221_3, 1222_3, 2000_3, 2001_3, 2002_3, 2010_3, 2011_3, 2012_3, 2020_3, 2021_3, 2022_3, 2100_3, 2101_3, 2102_3, 2110_3, 2111_3, 2112_3, 2120_3, 2121_3, 2122_3, 2200_3, 2201_3, 2202_3, 2210_3, 2211_3, 2212_3, 2220_3, 2221_3, 2222_3, 10000_3, 10001_3, 10002_3, 10010_3, 10011_3, 10012_3, 10020_3, 10021_3, 10022_3, 10100_3, 10101_3, 10102_3, 10110_3, 10111_3, 10112_3, 10120_3, 10121_3, 10122_3, 10200_3, 10201_3, 10202_3, 10210_3, 10211_3, 10212_3, 10220_3, 10221_3, 10222_3, 11000_3, 11001_3, 11002_3, 11010_3, 11011_3, 11012_3, 11020_3, 11021_3, 11022_3, 11100_3, 11101_3, 11102_3, 11110_3, 11111_3, 11112_3, 11120_3, 11121_3, 11122_3, 11200_3, 11201_3, 11202_3, 11210_3, 11211_3, 11212_3, 11220_3, 11221_3, 11222_3, 12000_3, 12001_3, 12002_3, 12010_3, 12011_3, 12012_3, 12020_3, 12021_3, 12022_3, 12100_3, 12101_3, 12102_3, 12110_3, 12111_3, 12112_3, 12120_3, 12121_3, 12122_3, 12200_3, 12201_3, 12202_3, 12210_3, 12211_3, 12212_3, 12220_3, 12221_3, 12222_3, 20000_3, 20001_3, 20002_3, 20010_3, 20011_3, 20012_3, 20020_3, 20021_3, 20022_3, 20100_3, 20101_3, 20102_3, 20110_3, 20111_3, 20112_3, 20120_3, 20121_3, 20122_3, 20200_3, 20201_3, 20202_3, 20210_3, 20211_3, 20212_3, 20220_3, 20221_3, 20222_3, 21000_3, 21001_3, 21002_3, 21010_3, 21011_3, 21012_3, 21020_3, 21021_3, 21022_3, 21100_3, 21101_3, 21102_3, 21110_3, 21111_3, 21112_3, 21120_3, 21121_3, 21122_3, 21200_3, 21201_3, 21202_3, 21210_3, 21211_3, 21212_3, 21220_3, 21221_3, 21222_3, 22000_3, 22001_3, 22002_3, 22010_3, 22011_3, 22012_3, 22020_3, 22021_3, 22022_3, 22100_3, 22101_3, 22102_3, 22110_3, 22111_3, 22112_3, 22120_3, 22121_3, 22122_3, 22200_3, 22201_3, 22202_3, 22210_3, 22211_3, 22212_3, 22220_3, 22221_3, 22222_3, 100000_3}

x² + 30x + 180  =  -36

                                         Add  36  to both sides of the equation.

x² + 30x + 216  =  0

                                         What two numbers add to  30  and multiply to 216 ?  +12  and  +18

                                         So the left side of the equation factors like this...

(x + 12)(x + 18)  =  0

                                         Set each factor equal to zero and solve for  x .

x + 12  =  0          or          x + 18  =  0

  x  =  -12                            x  =  -18

The roots of the equation are  -12  and  -18 .

The nonnegative difference between  -12  and  -18  is   (-12) - (-18)   =   -12 + 18   =   6

In right triangle $ABC$,
we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$.
If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is an altitude of $\triangle ABC$,
then what is $\cos \angle ABM$?

\(\begin{array}{|rcll|} \hline \angle \text{ABM} &=& 90^{\circ}-\angle \text{CAB} \\ \cos(\angle\text{ABM}) &=& \cos(90^{\circ}-\angle \text{CAB}) \\ \cos(\angle \text{ABM}) &=& \sin(\angle \text{CAB}) \quad & | \quad \sin(\angle \text{CAB}) = \dfrac{BC}{\sqrt{AB^2+BC^2} } \\ \cos(\angle \text{ABM}) &=& \dfrac{BC}{\sqrt{AB^2+BC^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{10^2+24^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{676} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{ 26 } \\\\ \mathbf{ \cos(\angle \text{ABM}) } & \mathbf{=} & \mathbf{ \dfrac{12}{ 13 } } \\ \hline \end{array} \)

What is the sum of all positive integers v for which lcm(v,20)=60.

\(\begin{array}{|r|r|r|r|c|} \hline v & \text{factor } v &\text{factor } 20 & \text{factor } 60 &\text{lcm}(v,20 ) \\ \hline 3 & {\color{red}3} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 6 & {\color{red}3} \times 2 & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 15 & {\color{red}3} \times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 30 & {\color{red}3} \times 2\times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 12 & {\color{red}3} \times {\color{blue}2^2} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 60 & {\color{red}3} \times {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline \text{sum } = 126 \\ \hline \end{array} \)

Let $n$ equal the number of sides in a regular polygon.
For \( 3\leq n <10\),
how many values of $n$ result in a regular polygon
where the common degree measure of the interior angles is not an integer?

Formula:

\(\begin{array}{lcll} \text{interior angle $= 180^{\circ} - $ exterior angle $\\$ }\\ \text{exterior angle $= \dfrac{360^{\circ} }{n} \\$ }\\ \boxed{\text{interior angle}= 180^{\circ} - \dfrac{360^{\circ} }{n} }\\ \end{array} \)

Note that an integer value for an interior angle's measure will occur  whenever  n  divides \(360^{\circ} \) evenly

The divisors of 360 are:

\(\text{$1 | 2 | {\color{red}3} | {\color{red}4} | {\color{red}5} | {\color{red}6} | {\color{red}8} | {\color{red}9} | 10 | 12 | 15 | 18 | 20 | 24 | 30 | 36 | 40 | 45 | 60 | 72 | 90 | 120 | 180 | 360 (24 $ divisors ) }\)

Only Number 7 is not a divisor of 360,

so only the polygon with a side of 7 will not have an interior angle with an integer-valued measure

If loga(s) = 5 and loga(t) = 10, then loga(s/t2) = ?

\(\begin{array}{|rcll|} \hline \log_a(s) &=& 5 \\ \log_a(t) &=& 10 \\\\ \log_a(s) - 2\cdot \log_a(t) &=& 5 - 2\cdot 10 \\ \log_a(s) - 2\cdot \log_a(t) &=& -15 \\ \log_a(s) -\log_a(t^2) &=& -15 \\\\ \mathbf{\log_a\left(\dfrac{s}{t^2} \right)} & \mathbf{=} & \mathbf{-15} \\ \hline \end{array}\)

Any value of  y  in the interval  (10, 11)  will make the equation true, for example....

\(\lceil{10.1}\rceil\cdot\lfloor{10.1}\rfloor\,=\,11\cdot10\,=\,110 \)

But also, any value of  y  in the interval  (-11, -10)  will make the equation true, for example...

\( \lceil{-10.9}\rceil\cdot\lfloor{-10.9}\rfloor\,=\,-10\cdot-11\,=\,110\)

Since the problem says  y < 0 ,  the only values of  y  that work are those in the interval  (-11, -10) .