hectictar

(b)

We start with  sqrt(x)  on the display.

Then take the  arctan  of what is on the display.

Then take the  cos  of what is on the display.  This gives us  $\frac{1}{\sqrt{x+1}}$

Then...just like from part (a).....we can take the  arctan  of what is on the display.

Then take the  cot  of what is on the display.

Altogether:

$\begin{array}{ccc} &\cot( \arctan( &\cos( \arctan(\sqrt{x}\,)\,) & ) \quad ) \\~\\ =&\cot( \arctan( &\frac{1}{\sqrt{x+1}}& ) \quad ) \\~\\ =&\sqrt{x+1} \end{array}$

I agree with EP. The locus would be the four points of intersection of the circle and the blue lines in this graph:

You're exactly right!! 👍 🤪  🙌

And you can always check your answer to these type of questions with a graphing calculator such as desmos.

$2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2x+2y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2(x+y)}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,(6^2)^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,36^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,36^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ (2\,\cdot\,36)^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ 72^{x+y}\,\cdot\,3^{x-y}\ =\ 72$

At this point we can see that the equation will definitely be satisfied if

x + y  =  1    and

x – y  =  0

So    x  =  1/2    and    y = 1/2    is definitely a solution.

But that doesn't necessarily mean it is the only solution.

I don't know how to prove that it is the only rational solution, but it does seem to be.

Maybe someone else can explain why that is.

Draw a line segment from  A  perpendicular to  DC  that meets  DC  at point  E , and

draw a line segment from  B  perpendicular to  DC  that meets  DC  at point  F , like this:

The sum of the interior angles in  △ADE  =  180°

45°  +  90°  +  m∠DAE   =   180°

m∠DAE   =   180° - 45° - 90°

m∠DAE   =   45°

△ADE  is an isosceles triangle and the sides opposite the base angles are congruent.

DE  =  AE

And if   DE = b   then   AE = b

By the Pythagorean Theorem,

DE² + AE²  =  (8√2)²

b² + b²  =  (8√2)²

2b²   =   128

b²   =   64

b   =   8

ABFE  is a rectangle so

BF  =  AE  =  8

EF  =  AB  =  4

By the Pythagorean Theorem,

BF² + FC²  =  10²

8² + FC²  =  10²

FC²  =  10² - 8²

FC²  =  36

FC  =  6

Now we know:

AB  =  4

So we can say  base₁ =  4

DE  =  8   and   EF  =  4   and   FC  =  6

So we can say  base₂  =  8 + 4 + 6  =  18

AE  =  8

So we can say  height  =  8

area of trapezoid  =  (1/2)(base₁ + base₂)(height)

area of trapezoid  =  (1/2)(4 + 18)(8)

area of trapezoid  =  88

If you have the 45-45-90 triangle memorized, you can immediately recognize that DE and AE  must be 8,

but if not you can always figure it out this way. 

$\phantom{=\quad}\sqrt{18\cdot n\cdot34}\\~\\ {=\quad}\sqrt{3\cdot3\cdot2\cdot n\cdot2\cdot17}\\~\\ {=\quad}\sqrt{3\cdot3\cdot2\cdot 2\cdot17\cdot n}\\~\\ {=\quad}\sqrt{3\cdot3}\cdot\sqrt{2\cdot 2}\cdot\sqrt{17\cdot n}\\~\\ {=\quad}3\cdot2\cdot\sqrt{17\cdot n}\\~\\ {=\quad}6\sqrt{17n}$

There must be a pair of  17's  in the prime factorization of the number under the sqrt.

So  17  must be a factor of  n  , and the least possible positive integer value of  n  is  17

$\phantom{=\quad}6\sqrt{17\cdot17}\\~\\ {=\quad}6\cdot17\\~\\ {=\quad}102$_

There are  5  choices for the first letter, then 26 choices for the second letter, then 26 choices for the third letter.

so total number of possible words   =   5 * 26 * 26   =   3380

What interval consists of all $w$ which satisfy $\bf neither$ $-2(6+2w)\le -16$ nor $-3w\ge 18$ ?

If  w  makes the inequality   -2(6 + 2w) ≤ -16   false, then

it must make the inequality  -2(6 + 2w) > -16   true.

-2(6 + 2w)  >  -16

                             Divide both sides by  -2, a negative number, so flip the sign

6 + 2w  <  8

                             Subtract  6  from both sides of the inequality.

2w  <  2

                             Divide both sides by  2, a positive number, so don't flip the sign

w  <  1

If   -2(6 + 2w) > -16   then   w < 1

If  w  makes the inequality   -3w ≥ 18   false, then

it must make the inequality  -3w < 18   true.

-3w  <  18

                  Divide both sides by  3, a negative number, so flip the sign.

w  >  -6

If   -3w < 18   then   w > -6

So the solution to the question is all  w  such that   w < 1   and   w > -6

That is all  w  in the interval   (-6, 1)

If we connect the centers of the three smaller circles, we get an equilateral triangle because the length of each side is twice the radius of a smaller circle. Then if we connect the center of the equilateral triangle to each of its three vertices, we get three new congruent triangles. Since these new triangles are congruent to each other,

m∠CBD  =  (1/2)(60°)   =   30°

And let's say

radius of smaller circle  =  AB  =  BD  =  r

And so  △CBD  is a  30-60-90  triangle where

the side across from the  60°  angle  =  r

the side across from the  30°  angle  =  r / √3

the side across from the  90°  angle  =  2r / √3

BC  =  2r / √3

Because the radius of the bigger circle  =  1

AB + BC  =  1

r  +  2r / √3   =  1

r( 1 + 2/√3 )  =  1

r   =   1 / ( 1 + 2/√3 )

r   =   2√3 - 3

Thanks guest, that is reassuring!!