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In acute triangle \(ABC,\ \angle A = 68^\circ\).

Let be the circumcenter of triangle \(ABC\).
Find, \(\angle OBC\) in degrees.

\(\text{Let $\angle OBC = {\color{red}x}$ } \)

\(\begin{array}{|rcll|} \hline 2{\color{red}x} + 2{\color{blue}y} + 2{\color{green}z} &=& 180^{\circ} \quad | \quad : 2 \\ {\color{red}x} + {\color{blue}y} + {\color{green}z} &=& 90^{\circ} \quad | \quad {\color{blue}y} + {\color{green}z} = \angle A \\ {\color{red}x} + A &=& 90^{\circ} \\ {\color{red}x} &=& 90^{\circ} - A \quad | \quad A = 68^{\circ} \\ {\color{red}x} &=& 90^{\circ} - 68^{\circ} \\ {\color{red}x} &=& 22^{\circ} \\ \hline \end{array}\)

\(\text{$\angle OBC$ in degrees is $22^{\circ}$}\)

Why just  a + b + c + d ? , isn't e a coefficient as well ?

Enter each fraction in the fields provided (formatted like 1/3, 2 7/8, or 7/5).

Select the operation you wish to perform (i.e. Add, Subtract, Multiple, or Divide).

Finally, click on the "Calculate" button to see the result.

Alan’s observation not only demonstrates a logical solution for this question, but also demonstrates how to think outside the box or the grid. When a process works, it’s very human to stay in our “space” and not consider ideas outside that space. Education, training, and experience all contribute to our thinking processes, and create predetermine modes of operational thought.

When we use our minds, instead of our machines, we see parts of our world become as new.

sin ^ 2x-9cos ^ 2x If cos = -0.9

\(find\;\;\;sin ^ 2x-9cos ^ 2x\;\;\;\;\; If \;\;\;\;cosx = -0.9\\~\\ sin ^ 2x-9cos ^ 2x\\ =sin ^ 2x-9cos ^ 2x+cos^2x-cos^2x\\ =(sin ^ 2x+cos^2x)-10cos ^ 2x\\ =1-10cos ^ 2x\\ =1-10*0.81\\ =1-8.1\\ =-7.1\)

The Stringbulider sub may have a string concatenation error.  There may be “hidden” data in the file that is read into the string. Generally a read-buffer reads numeric data as numeric and this includes an implied “+” sign which is not visible. This is common to most all programming languages

The error that occurs at “101” may also have occurred at “10,” but it didn’t display because it only displays if the number is palindromic. Temporarily set the output to screen-print all converted numbers to see if this is the case.  If it is, then step through the logic to find and correct the length error. 

The error at 227 is probably cause by rubbish in the input file. There may be a trailing decimal following the number. This will cause the byte size to jump by 2, because the input read buffer will treat it as a floating point number, instead of an integer.

The fix for these errors is usually trivial. It may require stepping through the logic, examining every parameter, checking for string length, and for residual data that may remain in a buffer.   

GA

help
Let

\(\large{f(x) = \dfrac{2x + 3}{kx - 2}}\).

Find all real numbers

\(\mathbf{k}\)

so that

\(\large{f^{-1}(x) = f(x)}\).

discontinuity:

\(\begin{array}{|rcll|} \hline 2x+3 &=& 0 \\ 2x &=& -3 \\ x &=& -\dfrac{3}{2} \\ \hline kx - 2 &\ne& 0 \\ kx &\ne& 2 \\ k &\ne& \dfrac{2}{x} \quad | \quad x = -\dfrac{3}{2} \\\\ k &\ne& \dfrac{2}{-\dfrac{3}{2}} \\\\ \mathbf{k} & \mathbf{\ne} & \mathbf{-\dfrac{4}{3}} \\ \hline \end{array} \)

At \(k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}\)

help

Let 

\(\large{S = \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb}\)

and
\(\large{T = \dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb}\).
Find

\(\large{\dfrac{S}{T}}\)

1.

\(\mathbf{S+T =\ ?}\)

\(\begin{array}{|rcll|} \hline S+T &=& \left(\dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb \right) +\left(\dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb \right) \\ S+T &=& \color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3} + \dfrac{1}{5^3} + \dfrac{1}{6^3} + \dotsb} \\ \hline \end{array}\)

2.

\(\mathbf{\dfrac{S}{T} =\ ?} \)

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3}+ \dfrac{1}{8^3}+ \dfrac{1}{10^3} + \dotsb \\ &=& \dfrac{1}{(2\cdot 1)^3} + \dfrac{1}{(2\cdot 2)^3} + \dfrac{1}{(2\cdot 3)^3}+ \dfrac{1}{(2\cdot 4)^3}+ \dfrac{1}{(2\cdot 5)^3} + \dotsb \\ &=& \dfrac{1}{2^3 1^3} + \dfrac{1}{2^3 2^3} + \dfrac{1}{2^3 3^3}+ \dfrac{1}{2^3 4^3}+ \dfrac{1}{2^3 5^3} + \dotsb \\ &=& \dfrac{1}{2^3} \underbrace{ \left(\color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3}+ \dfrac{1}{4^3}+ \dfrac{1}{5^3} + \dotsb} \right) }_{=S+T} \\ &=& \dfrac{1}{2^3}(S+T) \\\\ S &=& \dfrac{1}{8}(S+T) \quad | \quad \cdot 8 \\ 8S &=& S+T \quad | \quad -S \\ 7S &=& T \quad | \quad :7 \\ S &=& \dfrac{T}{7} \quad | \quad:T \\ \mathbf{\dfrac{S}{T}} & \mathbf{=} & \mathbf{\dfrac{1}{7}} \\ \hline \end{array}\)

I am not sure what you problems is. I have graphed them here

What is the area of the composite figure whose vertices have the following coordinates?

(−2, −2) , (4, −2) , (5, 1) , (2, 3) , (−1, 1)

\(\begin{array}{|r|r|r|r|r|} \hline \text{Point} & x & y & \\ \hline 1 & -2 & -2 &\\ & & & (-2)(-2) - 4(-2) & = 12 \\ 2 & 4 & -2 & \\ & & & 4\cdot 1 - 5 (-2) & = 14 \\ 3 & 5 & 1 &\\ & & & 5\cdot 3 - 2 (1) & = 13 \\ 4 & 2 & 3 &\\ & & & 2\cdot 1 - (-1)3 & = 5 \\ 5 & -1 & 1 &\\ & & & (-1)(-2) - (-2)1 & = 4 \\ 1 & -2 & -2 & \\ \hline & & & & \text{sum} = 48 \\ & & & & \text{area of the composite figure } = \dfrac{\text{sum}}{2} = \dfrac{\text{48}}{2} = \mathbf{24} \\ \hline \end{array}\)

\(f(x)=ax^4+bx^3+cx^2+dx+e\\ e=47\\ f(x)=ax^4+bx^3+cx^2+dx+47\\ f(1)=a+b+c+d+47=32\\ a+b+c+d=32-47\\ a+b+c+d=-15\\\)

Ths sum of the coefficients is -15

This would be true even if it the leading power was not 4.

\(A x = y\\ Ay = x + 2y\)

\(A^2x =AAx = Ay = x+2y\)

\(A^3x=AA^2x = A(x+2y) = \\ Ax + 2Ay = y + 2(x+2y) = \\ 2x+5y\)

\(A^4x = A(2x+5y) = 2y + 5(x+2y) = 5x+12y\)

\(A^5 x = A(5x+12y) = 5y + 12(x+2y) = 12x + 29y\)

\((a,b) = (12, 29)\)

Looks like 

OOEO and OEOO

both give 7 odds in the triangle

So it seems like the only way to solve this is with brute force?

Oh,  I'm pretty sure they can be though :/

We're assuming that a, b  are not the same integers....( I guess!! )

48 can be written with a=6 and b=6, as 6(6+1)+6=48, though

See my answer...I believe that it is correct...

Sorry, but my question was numbers that CANNOT be written as such, but thank you anyways. :)

(50 can be written as 2(16+1)+16=50 as my progam found.)

The fraction will not repeat if we can make a reduced fraction with only some power of "2" in the denominator

So

1/12   =  repeats

2/12 = 1/6  =  repeats

3/12 = 1/4  = 1/2^2   = terminates

4/12 = 1/3  = repeats

5/12 = repeats

6/12 = 1/2  = terminates

7/12 = repeats

8/12 = 2/3  = repeats

9/12 = 3/4 = 3/2^2 = terminates

10/12 = 5/6  = repeats

11/12 = repeats

I found that 46 is the largest number using python, b ut I don't know how to do it mathematically.

As long as the triangle passes the triangle inequality test

Then if

a^2 + b^2 = c^2       right triangle

But if 

a^2 + b^2  > c^2     then the triangle is acute

And if 

a^2 + b^2  <  c^2    then the triangle is obtuse