1 Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3 The ratio of lead nitrate to sodium iodide is 1/2
x/10 = 1/2 x = 5 moles
See below:
Hi Alan,
wow. That is the solution!
I followed your point 1 and finally found my answer :)
I'm so happy...
Thank you, from the depth of my heart!
Your karma has just risen by 1000 points minimum.
Keep up the good work here, you're
helping real people solve real technical problems.
Kind regards,
Robert
why should you get extra credit for asking other people to do the work?
Seems to me that an F would be more appropriate.
It depends on many things.
Your pre knowledge
Your ability
Your memory
etc.
For me, I think it was my memory that let me down the most.
I could understand fairly easily i guess but I had to do hours of repetitious exerceises almost every night in order to cement the techniques/ methods/ formulas etc into my brain.
If you do not understand the pre-learning then a good private tutor could help enormously.
A,B and C represent the areas
\(\frac{C}{A+B+C}=\frac{2}{5}\qquad (1)\\~\\ B=\frac{3}{7}(A+B)\\ \frac{7B-3C}{3}=A \qquad (2)\\ sub \;\;2\;\; into \;\;1 \\ \frac{C}{\frac{7B-3C}{3}+B+C}=\frac{2}{5}\\ \frac{3C}{7B-3C+3B+3C}=\frac{2}{5}\\ \frac{3C}{10B}=\frac{2}{5}\\ \frac{C}{B}=\frac{20}{15}\\ \frac{C}{B}=\frac{4}{3}\\ \frac{B}{C}=\frac{3}{4}\\\)
LaTex:
\frac{C}{A+B+C}=\frac{2}{5}\qquad (1)\\~\\ B=\frac{3}{7}(A+B)\\ \frac{7B-3C}{3}=A \qquad (2)\\ sub \;\;2\;\; into \;\;1 \\ \frac{C}{\frac{7B-3C}{3}+B+C}=\frac{2}{5}\\ \frac{3C}{7B-3C+3B+3C}=\frac{2}{5}\\ \frac{3C}{10B}=\frac{2}{5}\\ \frac{C}{B}=\frac{20}{15}\\ \frac{C}{B}=\frac{4}{3}\\ \frac{B}{C}=\frac{3}{4}\\
1. Draw a graph of the left-hand side against w and the right-had side against w to see where they intersect (perhaps multiply both sides by cos(w) first so that you don't get any divide by zeros).
2. Use a numerical technique like Newton-Raphson (though it's still a good idea to draw a graph first).
I m not sure what I am meant to do to show this.
To me it is just obvious.
Anyway ......if you let
\(z_1=a(cos\theta +isin\theta )\\ z_2=a(cos\alpha +isin\alpha )\\ \text{midpoint will be average of the real parts + average of the imaginary part}\\ midpoint=\frac{a(cos\theta +\cos\alpha)}{2}+\frac{a(sin\theta +\sin\alpha)}{2}\\ midpoint=\frac{a(cos\theta +\cos\alpha)+a(sin\theta +\sin\alpha)}{2}\\ midpoint=\frac{z_1+z_2}{2}\\\)
3 divided by 4 3/4 times
It seems that 2 in 8 are red
so
\(\frac{x}{12}=\frac{2}{8}\\ \frac{x}{12}=\frac{1}{4}\\ x=\frac{12}{4}\\ x=3\)
Based on experiemntal probability, 3 are likely to be red.
Yes, not big deal.
I liked your approach :)
Why didn't you use the calculator on your device
Idk if you are putting the balls back, but if you are, the answer would be 3
Khan academy is a good place to start reviewing concepts you don't understand.
bruh thats easy 75% of 100 how much is 100 a 100 and 75 pecent of 100 is???
75
Thank you!
It depends on your level.
I typically use Art of Problem Solving to practice math by playing Alcumus and watching videos every day. You just have to be focused on your goal of getting better at math.
YOU'RE SO WRONG, YOU'RE SO WRONG,
143, 209, 403, 407, 493, 667, 899, 961, 1147, 1271, 1457, 1961, 1963, 1969, 1991, 2021, 2059, 2071, 2077, 2101, 2201, 2209, NOT PRIME!!!
I'm sorry, i just need the answers to these question really fast, these were on one of my extra credit activities and my report cards are coming out on friday, these are the 4 questions that I got wrong and my teacher let me correct them. That is why I am posting all this in one thread.
Please do not post multiple questions in one thread.
After pouring off 1/4 of the fluid there is 3/4 left <=== half of this is Sprite or 3/8 ths
then she adds the 1/4 1/4 + 3/8 = 5/8ths is now soda
Thanks, Melody ... I misread it ....
