hectictar

By the SAS congruence theorem,  △PQT ≅ △SRT  So...

m∠QPT  =  m∠RST  =  37°

m∠QTP  =  180° - 50°   =   130°

m∠PQT  =  180° - 37° - 130°  =  13°

Since  QT = RT  we can substitute  QT  in for  RT in the next equation.

$RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP$

Now we can substitute  11 - TP  in for  QT in the next equation.

By the Law of Sines,

$\frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}$

Now that we know the length of  TP ,  we can find the length of  QT.

$QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}$

Finally, we can use the Law of Sines again to find  QP.

$\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}$

So we have found:

$\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}$

And all lengths are in meters.

If you want to know how much you'd have to add to  340⁴  to get  680⁴, we can solve this equation for  x:

x + 340⁴  =  680⁴

x  =  680⁴ - 340⁴

x  =  200450400000

So

200450400000 + 340⁴  =  680⁴

In other words,  680⁴  is  200450400000  more than  340⁴

If you want to know what power must  340⁴  be raised to to equal  680⁴,  we can solve this equation for  x:

(340⁴)^x   =   680⁴

340^x  =  680

x  ≈  1.1189

So

(340⁴)^1.1189  ≈  680⁴

680⁴   =   (2 * 340)⁴

680⁴   =   2⁴ * 340⁴

680⁴   =   16 * 340⁴

In other words,  680⁴  is  16  times  340⁴

Here is something taken from what I wrote in my notes on my phone a long time ago...

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a + b  =  10a + b   Divide both sides by  a

a/a + b/a   =   10a/a + b/a

1 + b/a  =  10 + b/a    Subtract  b/a  from both sides

1  =  10

So you'd be like, "NO SOLUTION!!!" right??? WRONG!! because  a  =  0  IS a solution to the original equation, but when you divided both sides by  a  it made it look like there was no solution. So that is why you gotta always say   a ≠ 0  whenever you divide by an unknown number. So then the final answer would be basically saying "No solution when  a ≠  0.   So  a = 0  might be a solution might not be I dunno." Then you gotta check if  a = 0  is a solution.

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So that is why   0  is missing from your list. 

When  you divided both sides by  b,  you didn't note that the following equation holds only when  b ≠ 0

Otherwise, it is true that  ab = 36

a + ab²   =   40b

a - ab²   =   -32b

The purple values are equal to each other and the blue values are equal to each other.

a - ab²   =   -32b                     Add  40b  to both sides of the equation.

a - ab² + 40b   =   -32b + 40b        Since  a + ab²  =  40b   we can substitute  a + ab²  in for  40b

a - ab² + a + ab²   =   -32b + 40b      The elimination method is really just like substitution 

a - ab² + a + ab²   =   -32b + 40b      Simplify both sides by combining like terms.

2a   =   8b          Divide both sides of the equation by  8

$\frac14$a  =  b

Now we can substitute this value for  b  into one of the original equations.

a + ab²  =  40b

                                    Substitute   $\frac14$a   in for   b

a + a($\frac14$a)²  =  40($\frac14$a)

                                    Simplify both sides of the equation.

a + $\frac{1}{16}$a³   =   10a

                                    Multiply through by  16

16a + a³  =  160a

                                    Subtract  16a  from both sides and subtract  a³  from both sides

0  =  144a - a³

                                    Factor  a  out of both terms on the right side

0  =  a( 144 - a² )

                                           Factor   144 - a²   as a difference of squares

0  =  a( 12 - a )( 12 + a )

                                           Set each factor equal to  0  and solve for  a

x  can be all real numbers except those that make  $\lfloor x^2 - 7x + 13\rfloor = 0$

...which means...

x  can be all real numbers except those that make  $0\ \leq\ x^2-7x+13\ <\ 1$

To find the  x  values that make the inequality true, it helps to look at a graph of  y = x² - 7x + 13:

We can see that there are no  x  values which make  y  equal  0 , just like Davis found.

But we can also see that there are some  x  values which make  y  fall between  0  and  1

Let's find what  x  values make  y  equal to  1

1  =  x² - 7x + 13

                                               Subtract  1  from both sides of the equation.

0  =  x² - 7x + 12

                                               Factor the right side of the equation.

0   =  (x - 3)(x - 4)

                                               Set each factor equal to zero and solve for  x

x - 3  =  0     or     x - 4  =  0

  x  =  3        or       x  =  4

Now we can see that if  x  is between  3  and  4,  y  will be between  0  and  1

So the domain is all real numbers except those exclusively between  3  and  4

In interval notation, the domain is  $(-\infty,3)\cup(4,\infty)$_

In order from oldest to newest:

You must have mistyped your question because the cosine of nothing is bigger than  1

Did you mean to say   cos θ = √3 / 3   ?     That is,   $\cos\theta\ =\ \frac{\sqrt3}{3}$   ?

(c)

Let's define functions  a,  b,  and  c  like this:

a(x)  =  cot( arctan(x) )   =   $\frac1x$       where  x  is any real number except  0

b(x)  =  cos( arctan(x) )  =  $\sqrt{\frac{1}{x^2+1}}$       where  x  is any real number

c(x)  =  cos( arcsin(x) )   =   $\sqrt{1-x^2}$       where  x  is a real number the interval  [-1, 1]

We start with  0  on the display.

Then we can take the  cos  to get  1

Then we can do  b  to get  $\sqrt{\frac12}$

Then we can do  b  to get  $\sqrt{\frac23}$

Then we can do  c  to get  $\sqrt{\frac13}$

Then we can do  b  to get  $\sqrt{\frac34}$

Then we can do  c  to get  $\sqrt{\frac14}$

Then we can do  b  to get  $\sqrt{\frac45}$

Then we can do  b  to get  $\sqrt{\frac59}$

Then we can do  a  to get  $\frac{3}{\sqrt5}$