CPhill

tanA + secA = e^x

tanA  = e^x - secA     square both sides

tan^2A  = e^[2x] - 2secA*[e^x] + sec^2A

tan^2A  = e^[2x] - 2secA*[e^x] + tan^2A + 1

0 = e^[2x] - 2secAe^x + 1

2secAe^x = e^[2x] + 1

secA  = (e^[2x] + 1 )/ (2e^[2x])

cosA  = (2e^[2x] ) / ( e^[2x] + 1 )

Note that Melody gets an extraneous solution of cosA = 0 ......but this cannot be so, because this would imply that secA  = 1/cosA  = 1/0     and this would be undefined in the original assumption

I assume that  f(x)  = x^2

So......f(-3) just means to sub in -3    and evaluate.......so we have

(-3)^2   = 9

Here's another approach........let the shorter side of the rectangle = x.......then the longer side = (7/5)x

So......the perimeter is given by :

2x + 2(7/5)x = 96     multiply through by 5

10x + 14x  = 480

24x = 480

x = 20     and (7/5)x = 28

So....the area = 20 * 28 =  560cm^2

7 * 50 minutes = 350 minutes     =   5 hours and 50 minutes   teaching

50 + 30   = 80 minutes  of lunch  = 1 hr and 20 minutes   lunch

So

350 + 80 =   430 minutes   =  7 hours and 10 minutes [ total time]

Note that  m< BRP  and m<CRQ = 90       

And m<BRP + m<BPR  = 90

Therefore, m<BPR  = m<CRQ

And m< PBR = m <QCR = 90

Therefore, by AA congruency.....ΔPBR is similar to ΔRCQ

Then

PB/BR  = RC/CQ      

4/ BR  = RC/18       but RC  = 2BR  ...so

4/BR  = 2BR/ 18    cross-multiply

4*18  = 2BR^2

72  = 2BR^2      divide both sides by 2

36 = BR^2          take the square root of both sides

6  = BR      so.........CR  = 2BR  = 2*6   = 12

And by the Pythagorean Theorem

PR^2 =  (PB^2 + BR^2) =  4^2 + 6^2  = 16 + 36   = 52

Likewise

RQ^2  = (RC^2 + CQ^2) = 12^2 + 18^2  =  144 + 324 = 468

So, again by the Pythagorean Theorem:

PQ^2  = RQ^2 + PR^2

PQ^2 =  468 + 52

PQ^2  = 520      take the square root of both sides

PQ = sqrt(520)  = 2sqrt(130)  = about 22.8

I don't see diagram, either....but each dial has 5 possible settings  [ if we allow for repeats]....so....

5 * 5 * 5   = 5^3   = 125 possible combinations

Mmmmm...I'm not sure about Guest's answer.....it could be correct......but consider

It will take Rick 18km/ [9k//hr]  = 2 hours to finish  

And runnining at his current pace, it will take D**k 2.5 hours to finish

But...he needs to cut this to 2 hours just to finish even with Rick

Call his normal rate per hour, R  and call his increased rate in order to catch Rick, (R + x)

And since R * T  = D.......we have this situation......

At his current rate, the distance he has left to run = R *2.5

And at  his his increased rate, the distance he has left to  run is the same....and this is given by:

(R + x)*2  = D     .....so.....equating the distance, we have......

R * 2.5  =  (R + x) * 2    simplify

2.5R  = 2R + 2x       subtract 2R from both sides

.5R  = 2x      divide both sides by 2

.25R   = x 

So.....he needs to run (R + .25R)    = 1.25R   = 5/4 his current rate just to tie Rick   

Consider this in light of the Guest's answer.......if 5/4  of D**k's current rate = 11.25km/hr, then his current rate must  just be 4/5 of this......but 4/5 of this = 9km/hr......the very same rate that Rick runs.....and if D**k runs at the same rate as Rick, he would be even with him, not 30 minutes behind....!!!

In my opinion.....we actually can't provide an exact numerical answer to this because we would have to know how far D**k has run and the time he has taken to do so to get his current rate.....

I may have blown a fuse, here......can some other mathematician look at  this   ????

We need to use polynomial long division here, Katie-Leigh

              3x^2   - 9x    +  27

x + 3    [  3x^3  + 0x^2  +  0x   -  4   ]

               3x^3  + 9x^2

              ______________________

                           -9x^2   + 0x

                           -9x^2  - 27x

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

                                        27x - 4

                                        27x + 81

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

                                               -85

So the result   is     [ 3x^2 - 9x + 27]  R  -85/(x + 3)

Sorry....the P[number is not 3 ]  =  5/6

This is more tedious than difficult......!!!......it's asking for the perimeter

Starting with the triangle on the right.....the two equal sides  = 2(2/sqrt(3))(1.5) cm  = 6/sqrt(3) =   2sqrt(3)  = A       

And we need to find the other side [ the base] for a calculation in the next triangle on the left....using the Law of Sines, we can calculate 1/2 the base length = x

[ x] / sin(30)  = 1.5 / sin(60)

2x  = [1.5 *2 ] / sqrt(3)

x =  1.5/[sqrt(3)]      and the base is twice this    3/sqrt(3)   = sqrt(3)

In the next triangle to the left, the missing angle  = 35 degrees    and we can find the base of this triangle using the Law of Sines

base / sin 90  =  [sqrt(3)] /sin(35)

So the base = [sqrt(3)]/sin(35)  =  B

And we  need to find the length of the remaining side of this triangle - the side with the two hash marks = for a calculation in the next triangle to the top left.......we'll call this missing side x, again

And with the Law of Sines, we have

x/ sin55 = [sqrt(3)]/ sin(35)    =    sin(55)*[sqrt(3)]/ sin(35) = C

And we can find the remaining top perimeter as

x/ sin(110)  = [sin(55)*[sqrt(3)]/ [sin(35)]/ [sin(35)]

So....the remaining top perimeter = [sin(110)][sin(55)*[sqrt(3)]/ [sin(35)]^2 = D

And we can find the leftmost side of the recatangle on the left as :

x /sin(35)  = sin(55)*[sqrt(3)]/ [sin(35)]

So this side  = sin(55)*[sqrt(3)]  = E

And the remaining length will be the bottom of the rectangle on the left......and this is just 1/2 the length of D  = (1/2)[sin(110)][sin(55)*[sqrt(3)]/ [sin(35)]^2  = F

So...the total perimemter=

A + B + C + D + E + F  =

[2sqrt(3) + [sqrt(3)]/sin(35) + sin(55)*[sqrt(3)]/ sin(35) + [sin(110)][sin(55)*[sqrt(3)]/ [sin(35)]^2 +

sin(55)*[sqrt(3)] + (1/2)[sin(110)][sin(55)*[sqrt(3)]/ [sin(35)]^2] cm

So the total perimeter  =  about 16.369 cm

P.S......you should check my math......this one was drawn out and I may have made a mistake someplace....!!!!