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I find it astonishing that this site in general would be specifically approved by a teacher and not the net in general.

I suspect web2.0calc.com, and probably some others sites, were specifically recommended by the teacher as a resource because it’s relatively obscure (known only to a few million) compared to Google and You Tube (known to billions). I doubt the teacher specifically forbids the use of Google and You Tube, because, as you say, they have brilliant teaching resources. They may not have been explicitly recommended because their use is so obvious. 

I do know one (now retired) high-level university mathematics professor who absolutely forbade the use of the net for researching solutions to math questions. The syllabus for his class stated clearly that any student violating this restriction will receive a failing grade and face expulsion. I LMAO after reading the syllabus: I thought for any given term where a university may offer this kind of advanced mathematics, there might be 30 to 40 students in all the universities around the world who could possibly understand this esoteric math enough to take such a class. Good luck in finding this information and analysis on the net.  LOL

As for stranger danger, this site is potentially a greater risk (because of the Private Messaging) compared to Google or You Tube. Fortunately, in the past eight years, I’ve only heard of two instances of criminal level harassment –both by the same person, on this forum. 

Generally, the biggest risk to students on this forum is contracting CDD –Contagious Dumbness Disease, (which they can get anywhere), and having a troll snarl and growl at them. This may be rude and off-putting to some, but it’s not criminal, and I do not throw the younger students off the bridge –only adults and older students that are obnoxious brats from heII.

This type of trolling is sometimes an effective method of reaching into the ever-expanding mind of a young student and disrupting the complacent, self-narrative that often gives rise to intellectual laziness. Ideally, the self-narrative will change enough to allow for a greater and accelerated assimilation of information and knowledge, allowing the student to reach a higher potential, faster.

Younger students usually run when I troll them. Elijah defended his position –I’m rather impressed. I think Elijah will be a wondrous student by the time he enters university.

Considering this, I think you should add this thread to the Sticky Topics rather than blanking the posts. (you can –or I can star-out the peppery words, if you wish).  I understand that this thread is a briar patch in a diverse garden. Large mammals and especially humans should not walk in such places without proper equipment. Briar patches have their own ecology; if you blank some or all of this, then this ecology may become extinct or perpetually dormant.  This seems like a waste. No one will die from reading this; the worse case is some will end their excursion with itchy pricks from the thorns. They will heal ...and life will continue.... 

Of course the final decision on its disposition is yours....

GA

--. .-

simplify the parenthesis, then solve the 3 step equation.

I WAS RIGHT >:( 

nice

Thanks for the help!

Oh! I see the mistake! Thanks for pointing it!
My solution is then completely wrong! Sorry!
Will try to find another solution.

For 3-digit integers, there are 5 of them as follows:

225 , 323 , 416 , 522 , 614 , Total =  5 such positive integers.

Yep, 

LaTex is really cool!   

But it was playing up for me today and not displaying properly  

Sorry, but I'm still not convinced.

At the point where you say  square both sides and solve the inequality , are you not multiplying throughout by negative quantities, ( -x on the rhs), in which case the direction of the inequality should switch.

-4 < -2, so squaring and solving the inequality, 16 < 4, 12 < 0 ?

It's okay, we were writing our responses at the same time.

Sorry Max, I didn't see your answer there. :)

Sorry, the Latex is playing up.

its (-1,1]

Sometimes I just gotta laugh.   

\(\frac{3-z}{z+1} \geq 1 \\ \text{ Assuming that z is a real number}\\ z\ne-1\\ \frac{3-z}{z+1} \geq 1 \\~\\ If\;\;z>-1\;\;then\\ 3-z \geq z+1 \\ -2z\ge -2\\ z\le1\\ \therefore -1

or

\(If\;\;z<-1\;\;then\\ 3-z \leq z+1 \\ -2z\le -2\\ z\ge1\\ \therefore no\;\;soln \)

Hence

\(-1

Ignoring the toxic "discussion" above and actually solving the problem:

\(\dfrac{3 - z}{z + 1} \geq 1\\ \dfrac{3 - z}{z + 1} (z + 1)^2 \geq (z + 1)^2\text{ and }z +1 \neq 0\\ (3 - z)(z + 1) \geq (z + 1)^2\text{ and }z +1 \neq 0\\ -z^2 + 2z + 3\geq z^2 + 2z + 1 \text{ and }z +1 \neq 0\\ 2z^2 - 2 \leq 0 \text{ and }z +1 \neq 0\\ z^2 - 1 \leq 0 \text{ and }z +1 \neq 0\\ -1 \leq z \leq 1 \text{ and }z \neq -1\\ -1 < z \leq 1\)

Alternative solution:

The sum of coefficients of a polynomial \(p(x)\) is actually just \(p(1)\).

Proof: (You can omit this part if you just want the answer instead of the explanation.)

Let \(p(x) = a_n x^n + a_{n - 1}x^{n - 1} + a_{n - 2}x^{n - 2} + \cdots + a_2 x^2 + a_1 x + a_0\).

Then \(p(1) = a_n 1^n + a_{n - 1} 1^{n- 1} + \cdots + a_1 \cdot 1 + a_0 = a_n + a_{n -1} + a_{n-2} + \cdots + a_1 + a_0\), which is exactly the sum of coefficients of p(x).

Therefore, we just find \(g(1) = f(1 + 7) = f(8)\). The sum of coefficients of g(x) is \(f(8) = 8^7 - 3\cdot 8^3 + 2\).

7 revolutions in 14 minutes means an average of 1 revolution per 2 minute.

That means the angular velocity of the Ferris wheel is \(\omega = \dfrac{2\pi}{120}\text{ rad s}^{-1} =\dfrac\pi{60} \text{ rad s}^{-1}\).

Using the formula \(v = r \omega\), we have the linear velocity = \(v = (30)\left(\dfrac\pi{60}\right) = \dfrac\pi2 \text{ ft s}^{-1}\).

Edit: That's 1.6 ft/s if you round it off to the nearest tenth.

Let P be the point on the tangent at the bottom of the image.

Note that \(\angle PXZ = \angle XZW\) since they are alternate angles of the pair of parallel lines.

Also, \(\angle PXZ = \angle XWY\) since they are angles in alternate segments.

Therefore \(\angle XZW = \angle XWY\). Also, \(\angle ZXW = \angle WXY\) since they are the same angle.

By AA postulate, \(\triangle XZW \sim \triangle XWY\).

Let YZ = t. Then by similar triangles, \(\dfrac{XZ}{XW} = \dfrac{XW}{XY}\). i.e., \(\dfrac{14 - t}{6} = \dfrac{6}{14}\).

Can you take it from here?

Since A + 2 = C + 2 = D + 2, A = C = D.

Since A + 2 = B - 2, B = A + 4.

Then A + (A + 4) + A + A = 36. i.e., A = 8.

Then A = C = D = 8 and B = 12. ABCD = 8^3 * 12 = 6144.

Hope this clarifies:

If latex is lagging: * 5x^2-204 4x^2-204 < 0 ----->x^2 < 51 -----> x <=7

for some reason latex does not work for 5x^2-204<0     

Oh! Sign error! Sorry!

But, anyway, a similar approach can be used:

Now the correct equation (2):

\(x^2-51=y(y+x)\)

The R.H.S is always positive so: \(x^2-51>0 \implies x > 7\implies x\ge8\)  (*) 

\(y=-\frac{x\pm \sqrt{5x^2-204}}{2}\)

\(y>0 \implies \frac{x\pm \sqrt{5x^2-204}}{2}<0\)

\(x\pm \sqrt{5x^2-204} <0\)

So: \(\pm \sqrt{5x^2-204}<-x\)  (Square both sides, and solve the inequality:)

Then: \(5x^2-204

\(4x^2<204\)

\(x^2<51\)

So: \(x\le 7\) (**)

But, (*) contradicts (**). (As there is no x that satisfies both)

Therefore, there does not exist an integer x so that y is a positive integer.

(1) T = 1/3 (C + D)

(2) C = 2/5 (T + D)

(3) T = D - 24

   D - 24 = 1/3 (C + D)

 3D - 72 + C + D

   C = 2D - 72

   C = 2/5 (D - 24 + D)

   C = 2/5 (2D - 24

 2D - 72 = 2/5 (2D - 24)

 5D - 180 = 2D - 24

 3D = 156

 Dave = $52

    T = 52 x 24

 Terry = $28

    C = 2 (52) - 72

 Carl = $32

The present cost = 52 + 28 + 32

                           = $112

LMAO, I realised my mistake!!!!

I should have looked at the final answer before submitting it!!!

-Vinculum

2r, = 10 yd

r1 = 5yd

r2 = r1 + 11yd

    = 5 + 11 yd

r2 = 16 yd

Area of Shaded region

          = 3.14r2²- 3.14r1²

          = 3.14 (r2²-r1²)

          = 3.14 (16² - 5²)

          = 3.14 (231)

          = 725.34 yd²