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The law of sines can be summerized pretty well in this picture I found on web (source: www.mathwarehouse.com). If you know an angle and its opposite length, you can solve for every other angle and length for that triangle. 

Ammonia is NH₃ 

The molar mass (reading from the periodic table) is 17.031g/mol

Therefore 3.4 grams of ammonia is equal to 0.1996359579590159 moles of ammonia.

Multiplying this by 6.022 * 10²³ we get 120220773882919374980000 molecules (or 1.2022 * 10²³ molecules).

In 1 molecule of ammonia there are 4 atoms, so in total there are approx. 4.8088 * 10²³ atoms.

The measurement was made using 2 significant figures, so we can only be sure about this result to 2 significant figures, therefore the amount of atoms in 3.4g of ammonia is approx. 4.8 * 10²³ atoms.

Find the largest negative integer x which satisfies the congruence

\( 34x+6\equiv 2\pmod {20}\).

\(x = -6+10n \quad | \quad n \in \mathbb{Z}\)

The largest negative integer x = - 6

Check:

\(\begin{array}{|rcll|} \hline && 34\cdot(-6)+6 \pmod {20} \\ &\equiv & -204 + 6 \pmod {20} \\ &\equiv & -198 \pmod {20} \\ &\equiv & -18 \pmod {20} \\ &\equiv & -18+20 \pmod {20} \\ &\equiv & 2 \pmod {20} \\ \hline \end{array}\)

Find the least positive integer x that satisfies 

\(x+4609 \equiv 2104 \pmod{12}\).

\(\begin{array}{|lrcll|} \hline & x+4609 &\equiv& 2104 \pmod{12} \\ & x+4609-2104 &=& n \cdot 12 \quad & | \quad n \in \mathbb{N} \\ & x+2505 &=& n \cdot 12 \\ & x&=& n \cdot 12 -2505 \\\\ \text{so } & n \cdot 12 &=& 2505 \\ & n &=& \uparrow \frac{2505}{12} \uparrow \\ & n &= & \uparrow 208.75 \uparrow \\ & n &= & 209 \\\\ & x&=& 209 \cdot 12 -2505 \\ & x&=& 2508 -2505 \\ & \mathbf{x} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

Hi Guest,

please try again. password reset page was updated

Admin

Hi Guest,

please try again. password reset page was updated

Admin

Hmmmm.... interesting...

There is only a certain number of posts that you're supposed to be able to make as a guest per day. I don't "work" here or anything.....I'm just a random user, haha. I have no control over getting your password back.

Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A.

If AB = 4,

   AD = 8,

and AC is perperpendicular to AD,

then find CD.

\(\begin{array}{|rcll|} \hline AD^2 &=& AB \cdot AC & \text{secant-tangent theorem} \\ \mathbf{AC} & \mathbf{=} & \mathbf{ \frac{AD^2}{AB} }\\\\ CD^2 &=& AD^2 + AC^2 & \text{pythagoras} \\ CD^2 &=& AD^2 + AC^2 & | \quad AC= \frac{AD^2}{AB} \\ CD^2 &=& AD^2 + \left(\frac{AD^2}{AB}\right)^2 \\ CD^2 &=& AD^2 + \frac{AD^4}{AB^2} \\ CD^2 &=& AD^2\cdot \left( 1+\frac{AD^2}{AB^2} \right) \\ CD^2 &=& AD^2\cdot \left[1+\left(\frac{AD}{AB}\right)^2 \right] \\ \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } }\\ \hline \end{array} \)

CD = ?

\(\begin{array}{|rcll|} \hline \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } } \quad | \quad AB=4 \quad AD=8 \\ CD & = & 8 \cdot \sqrt{1+\left(\frac{8}{4}\right)^2 } \\ CD & = & 8 \cdot \sqrt{1+ 2^2 } \\ CD & = & 8 \cdot \sqrt{5 } \\ CD & = & 17.88854382 \\ \hline \end{array}\)

I think the question you're talking about was deleted.

It isn't being answered on that page. Normal users can't see it.

Wasn't your question just about how check marks get there?

If so, I just answered the question in my last response.

Will i be able to see my answer after you finish answering it? Will you remove the red mark (not the check mark) that blocks me from reading it?

How can you not post new questions when you just posted this one? lol

In the graph below, each grid line counts as one unit.

The line shown below passes through the point  (not shown on graph).

Find n .

The line passes through the points (0,3)  and (-7,-1)

 \(\begin{array}{|rcll|} \hline \text{slope } = \frac{3-(-1)}{0-(-7)} &=& \frac{n-3}{1001-0} \\ \frac{3-(-1)}{0-(-7)} &=& \frac{n-3}{1001-0} \\ \frac{4}{7} &=& \frac{n-3}{1001} \\ 1001\cdot \frac{4}{7} &=& n-3 \\ 3+ 1001\cdot \frac{4}{7} &=& n \\ \mathbf{575} & \mathbf{=} & \mathbf{n} \\ \hline \end{array}\)

The blue line just means that a new post was made to that thread.

It only shows up on threads that you have looked at before.

And only mods have the power to check mark a question. A mod has to personally check every answer that gets check marked to make sure it is right and then give it the check mark.

No im not that guy. My question was about how and when you mark questions. 

Please answer my full question.

AND WHY IS THAT QUESTION MARKED WITH A NEW THING (that weird blue line)

I'm pretty sure your other question got deleted because you put a cuss word in it.

Are you the person who keeps asking about dropping 2 rocks and wanting to know info about the center of mass? If so, an answer was posted to it just recently.

Solution(s)

\(\text {(a) }\\ \text {Find coordinates (y) for masses } \displaystyle m_1(y_1) \text { and } m_2(y_2)\\ y_1 = 0.5(9.81)*(0.300)^2 = 0.441 \small \text{meters}\\ y_2 = 0.5(9.81)*(0.300 – 0.100)^2 = 0.196 \small \text{ meters}\\ \\ \small \text{ Formula for the center of mass (2 objects):}\\ \\ y_{com} = \dfrac{(m_1y_1 + m_2y_2)}{(m_1 + m_2)}\\ y_{com} = \dfrac{((1) (0.441) + (2)(0.196)}{(1) + (2)} = 0.278 \small \text{ meters} \leftarrow \small\color{green} \text{Answer for (a)}\\ \text {(b) }\\ \text{Velocity of center of mass}\\ v1=(g)(t)_1 = 9.8*0.300 = \small \text{ 2.94m/s }\\ v2=(g)(t)_2 = 9.8*0.200 = \small \text{ 1.96m /s }\\ v_{com} = \dfrac{((1)(2.94)+(2)(1.96))}{3} = 2.287 \small \text{ m/s} \leftarrow \small\color{green} \text{Answer for (b)}\\\\ \small \text{ }\\ \small \text{Theory & Formulas: Complements of Archimedes of Syracuse, Leonhard Euler, Sir Isaac Newton, Pappus of Alexandria,}\\ \small \text{Guido Ubaldi, Francesco Maurolico, Federico Commandino, Simon Stevin, Luca Valerio, Jean-Charles de la Faille, Paul Guldin,}\\ \small \text{John Wallis, Louis Carré, Pierre Varignon, and Alexis Clairaut.}\\ \small \text{ } \hspace{20em} \scriptsize \text {(Probably a few others, too. The COM concept is ancient) }\\ \small \text{Produced by Lancelot Link and company. }\\ \small \text{Directed by GingerAle. }\\ \small \text{Sponsored by Naus Corp. Quantum Pharmaceuticals. }\\ \small \text{ } \hspace{15em} \scriptsize \text { Making a better world by neutralizing the quantum dumbness of }\\ \small \text{ } \hspace{15em} \scriptsize \text { Blarney Masters and related dumb-dumbs }\\\)

Volume of cube  =  x^3

Volume of sphere  =   (4/3)pi* r^3

And

3(4/3)pi * r^3  =  x^3

4pi * r^3   =  x^3 

r^3  =  x^3 / [ 4pi]     take the cube root of both sides

r  =  x / (4pi)^(1/3)    → r^2  =  x^2 / (4pi)^(2/3)

Surface area of sphere  =

4pi * r^2  =  4pi * (x^2) / (4pi)^(2/3)  =  (4pi)^(1/3)*x^2 =  ∛[4pi] * x^2

perimeter of a rectangle = length + length + width + width

72 = 2w + 2w + w + w

72 = 6w

12 = w

width = 12 feet

length = 2 * w = 24 feet

If you are asking about the calc on this website, there are 2 bubbles at the very bottom left. Just select degrees or radians. I think it is set to degrees by default.

(On the home page it is set to degrees by default, but when I open it while in the forum it is set to radians...huh that's weird...)

2w + 3 < 11

2w < 11 - 3

2w < 8

w < 8 / 2

w < 4

There is a not so pretty drawing of the number line. :)