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OK, catmg   !!!

cot x - (csc^2 x / cot x)

cot^2 x / cot x = cot x

(cot^2 x / cot x) - (csc^2 x / cot x)

(cot^2 x - csc^2 x) / cot x

I learned that 1 + cot^2 x = csc^2 x today, thank you cphill. :)))

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cot x            cot x                 cot^2 x

____  *     _______  =       _______

  1               cot x                cot x

So

cot ^2x         csc^2 x

_____  -     _______  =  

cot x            cot x

(cot^2 x  -csc^2 x)

_________________

         cot x

OK.....glad to help .....

why does the cot^2 x happen after finding the common denominator?

You're welcome :)))

Thank you for correcting me and responding. 

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cot x  - ( csc^2 x  /cot x)                get a common denominator

cot ^2x   -csc^2  x

______________                 ( 1 + cot^2 x  = csc^2x   ⇒   cot^2 x   - csc^2 x  =  -1  )

    cot x

-1 /  cot x  =   

-1  / ( 1 /tan x) 

-1 (tan x)   =

-tan x

I would very much hope so! Thank you for your gracious attempt though!

oh noessss. :(((((

Welp, at least I tried. 

Luckily, it looks like CPhill is typing so hopefully we can learn from him. :))

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21.195 was the correct answer, but Thank you very much again!. I got it correct using 3.14 as pi not pi :)

We can split this into a rectangle and triangle. 

Rectangle

90*130 (length * width)

Triangle

110*50/2 (base * height / 2)

Now, we just add them together. 

90*130 + 110*50/2  

11700 + 2750

14450

I hope this helped. :))

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Not totally  sure about this one since oblique cylinders  don't  come along very often on here.....but  I  believe  that   we only need  the  radius and  height....the  axis  doesn't  matter

Using  this....the  volume  is  pi  * r^2 * h =   pi * (1.5)^2 * 3  = 

6.75 pi  yds^3    ≈  21.2 yds^3

I hate to be this person but this problem has to be solved whilst keeping -tan alone on the right side

thank you ((: 

Thank u

Being completly honest, I am not good with these things, but I'll give it my best shot. 

Let's start by turning everything into sine/cosine. 

(cos θ/sin θ) - (1/sin θ^2)/(cos θ/sin θ) = -(sine θ/cos θ)

(cos θ/sin θ) - (1/(sin θ * cos θ)) = -(sine θ/cos θ)

(cos θ^2 - 1)/(sin θ * cos θ) = -(sine θ/cos θ)

(cos θ^2 - 1) = (sin θ * cos θ) * -(sine θ/cos θ)

(cos θ^2 - 1) = - (sin θ^2)

cos θ = sqrt(1-sine θ^2)

This equation can be proven by using the unit circle and pythagreon theorum. 

Now, let's plug that into the original equation. 

(sqrt(1-sine θ^2)^2 - 1) = - (sin θ^2)

(1-sine θ^2 - 1) = - (sin θ^2)

- sine θ^2 = - (sin θ^2)

- sine θ^2 = - sine θ^2

omgosh, I can't believe I solved that. (hopefully it's correct)

I hope this helps. :)))

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'ft'   *   12   =  ' inches '

Hi

How many numbers are perfect squares. 

sqrt(100) = 10

sqrt(200) = 14.14...

So there are 5 squares from 100 to 200. 10^2, 11^2, ..., 14^2. 

From 100 to 200, there are 101 numbers, 5 of which are squares. 

101 - 5 = 96. 

I hope this helped. :))

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Triangle ADB  is similar to triangle  BDC

So

AD  / DB =  DB / DC

3  / DB  =  DB  /  12       cross-multiply

12 * 3  = DB * DB

36  = DB^2

DB  = 6

AB = sqrt (DB^2  + AD^2)  =   sqrt  ( 6^2  + 3^2)    =  sqrt  ( 45)  =  sqrt (9 ) * sqrt (5)  = 3sqrt (5)

You were correct, Thank You :)

If 1 in  = 1 ft

Then the  model must have been 196  in tall

V =  area of  base  *  height  /  3

area of  base =  product  of leg lengths of  the right triangle  /  2  =   10 * 12  /  2   = 60

So

V =   60  * 20  /  3   =   1200  /  3     =   400 cm^3

We  have  two triangles  each with a base of 7.7m  and a height of  7.0 m

Their total area = 7 * 7.7    = 53.9 m^2

And we have two rectangles with dimensions of 8 * 18

Their total area =  2 * 18 * 8   = 288 m^2

Finally  we  have a 7.7  by 18  rectangle  in the  center  =  7.7 * 18   = 138.6 m^2

Area =  [  53.9 + 288 + 138.6 ]  =   480.5 m^2

We  have  to follow  the  triangle inequality  that  says  that  the sum of  the lengths of any two sides of a triangle  is  greater  than  the remaining  side

B is  out   because   1 + 2   isn't greater  than 3

C is out  because  2 + 2  <  5

D is out because   2 + 6  <  9

A is correct

6 + 8 >  9

6 + 9 >  8

8 + 9 >  6

Here's  my  best  shot

If  the 10 ntegers  have  an average  of  6.....then  the  total sum of the  integers must  be 10 * 6   =  60 because  60 / 10  =  6

Therefore......we  want  to make  the  sum  of  the  "non-one" digits  as small as possible  and  the  number  of ones as large as possible

Possible Number of 1's         Sum  of  the other digits

              1                                      59

              2                                      58

              3                                      57

              4                                      56

              5                                      55

              6                                      54

              7                                      53

              8                                      52

              9                                      51

            10                                      50

The last is clealrly not possible  ( all ones)

The next-to-the last  is also  not possible......the  largest  sum  that  could be  created  with 9 ones  is  9 + 14 = 23

The  third-to-the last is also  not possible ...the largest possible sum  =  8 ones + 2 * 14  =  36

The fourth-to-the-last  is also not possible....the largest sum =  7 ones  + 3 * 14  = 49

Six  ones   is possible  because   14 + 14  + 13 + 13  =  54

So....the max number of 1s   =    6