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1)  " Triangle PQR~Triangle PUT, not similar. "

This says, "triangle PQR is similar to triangle PUT, not similar."

If they are not similar then we can't make a similarity statement!

But these triangles actually are similar...

∠PQR  ≅  ∠PUT   because they both are marked by a red line, and

∠QPR  ≅  ∠UPT   because they are vertical angles.

Since two of the angles are the same, the triangles are similar.

△PQR ~ △PUT

2)

∠VTU  ≅  ∠DTE  because they are vertical angles.

If these triangles are similar, then this must be true of the sides adjacent to the angle we know is shared by both triangles:

\(\frac{\text{shorter side of first triangle}}{\text{shorter side of second triangle}}\,=\,\frac{\text{longer side of first triangle}}{\text{longer side of second triangle}} \\~\\ \frac{19}{63}\,=\,\frac{27}{98} \)

Since this is not true, the triangles are not similar.

3)

If the triangles are similar, then this must be true:

\(\frac{32}{56}\,=\,\frac{40}{70}\,=\,\frac{48}{84} \\~\\ \frac47\,=\,\frac47\,=\,\frac47\)

Since this is true, the triangles are similar.

△EDC ~ △JKL

Ohhh I see, I'm sorry >.<

And

Angle C is opposite the longest side           Angle  C is opposite the longest side

Angle D is opposite the intermediate side   Angle  M is opposite the intermediate side

Angle E is opposite the shortest side         Angle L is oposite the shortest side 

So  Δ CDE  is similar to Δ CML

Look at this

Angle B  is opposite the longest side            Angle J is opposite the longest side

Angle C is oppositie the mid-length side       Angle K is opposite the mid-length side       

Angle D is opposite the shortest sde           Angle L is opposite the shortest side

So

 ΔBCD   is similar to  ΔJKL

Good job, CPhill!

Thank you all!!

Yes, that's correct.

√1/4 /  ⁴√1/8

Manuel....I find these easiest if we convert to exponential form, first...so we have

(1/4)^(1/2)  / ( 1/8)^(1/4)   note  that we can write

(1 / 2^2)^(1/2)  / (1 / 2^3)^(1/4)

( 1 / 2)  / ( 1/ 2^(3/4) )   ...so we have

(1/2)^1  *  (1/2)^(-3/4)  =

(1/2)^(1/4)   =  ⁴√(1/2)

If \(\underset{x\rightarrow0}{lim}f(x)=1\), we have not indefinite case \(\frac{0}{0}\), so we have not a real problem, and we can't use for example the l'Hospital rule . We must have \(\underset{x\rightarrow0}{lim}f(x)=0.\)

Original price 80 dollars

less 25%   80  -  80(.25) = 60 dollars     now take an additional 10% off

60- 60(.1) = $ 54.00

Price after off =80-  1/4 * 80 = 80-20 = $60

Coupon discount = 60*10/100 = $6

So, he will have to pay, 60-6 = $54

$80 x 1/4 =$20 - Discount on the jacket.

$80 - $20 = $60 - Sale price of the jacket

$60 - [10% x $60] =

$60 - [0.10 x $60] =

$60 -   $6.00 = $54 - The price of the jacket after the discount and 10% coupon.

Also, shouldn't CDE~MCL

Nevermind I solved it...

I think you got some of the triangles wrong. Shouldn't BCD~KJL?

Thank you! ^-^

First one top left....we can check these ratios

9/54  = 10/60  = 12/72   = 1/6   Similar due to SSS proportionality

So  ΔBCD  is similar  to  ΔJKL

Top  right

Check these ratios

60/156  = 65 /169 

5/13  =  5/13     Similar   due to SAS proportionality

So   ΔFGH is similar to  ΔSRQ

Bottom left

Check these ratios

40/96  = 65/156  = 70/168

5/12   = 5/12  = 5/12  Similar  due to SSS proportionality

So  ΔCDE is similar to  ΔCML

Bottom right

Not similar  -  no angles match

triangle BCD is similar to triangle KJL because of SSS similarity.

triangle FGH is not similar to triangle SRQ.

triangle CDE is not similar to triangle MLC.

triangle LMN is not similar to triangle FEG.

ln{[(5280×(236674+30303× sqrt(61)))³ +744]² - 393768} ÷ sqrt(1708) =?

ln{[(5280 x(236674+30303 x 7.810...))^3+744]^2-393768} / 41.32795... =

ln{[(5280 x (236674 + 236,673.995...))^3 +744]^2-393768} / 41.32795...=

ln{[(5280 x (473,347.995....)^3 + 744]^2 - 393768} / 41.32795 =

ln{[(2,499,277,418.505)^3 + 744]^2 - 393768} / 41.32795...=

ln{[15,611,455,512,523,783,919,812,598,068.719 + 744]^2 - 393768} / 41.32795....=

ln{[15,611,455,512,523,783,919,812,598,812.719]^2 -393768} /41.32795 =

ln{[243 717 543 219 509 240 868 385 073 078 387 627 457 317 726 561 954 328 436.172961]-393768} / 41.32795.....=

ln{243 717 543 219 509 240 868 385 073 078 387 627 457 317 726 561 954 328 436.172961} / 41.32795....=

129.8356049666 6784708192 4400512611....../ 41.32795 =

=3.141592653589793238462643383279502884197169399375105820974944592307816406286208..........!!!!!!!!!!

The answer is Pi accurate to 78 decimal places!!!!!!!!!!!!!!!!!!.

Note: I carried all computations to 100 decimal places where necessary!!.

Tysm you are seriously amazing! <3

(i know you meant UVW on the first one)

First one :

 ΔRQP   similar to  ΔUVW  by AA  similarity

For the second one :

Check these ratios

23/66 =  32/88  = 40/110

23/66 = 4/11 = 4/11   NOT similar

Since you are looking for a vertex form, I assume you are starting with a QUADRATIC:
-z^2 + 4z =4

-z^2 +4z-4 = 0         we want     a(z-h)^2 +k =0   form

-1 (z^2 -4z +4) = 0    'complete the square'

-1  {(z-2)^2 +4 -4} = 0

-1 (z-2)^2 +0 = 0             is the vertex form   where vertex is 2, 0

1.   (csc^2(x) - cot^2(x)) / (sin(-x)cot(x)

      Remember that  1 + cot^2 (x)  = csc^2 (x)

      So  csc^2 (x) - cot^2(x)  =  1

      Also   sin (-x)  =   - sin (x)

So  we have

1 / [ - sin (x) (cos (x)  / sin (x) ]  =

1 / (-cos (x) )  =    -  sec (x)

2.  sin ^2 (x)cos^2(x)  - cos^2(x)  =

(1  - cos^2(x) ) (cos^2 (x) ) -  cos^2(x)

cos^2(x) - cos^4 (x)   - cos^2(x)

-cos^4(x)

Tysm!! I have a ton more to do >.< 

1.  remember   cos^2 + sin^2 = 1

            sin^2 = 1- cos^2

             sin = sqrt (15/16)       (we are only concerned about the POSITIVE root in the 1st quadrant)

tan = sin/cos       sqrt(15/16) / 1/4   = 4 sqrt(15/16)   = 3.873

2.  sin (theta - pi/2)    results in shifting the sin wave pi/2 to the right which is an inverted cos wave

       so sin (theta-pi/2)    is   - cos(theta) =  - .54