hi please help me how to factor this out: 36x-45x^4+21x^3 thanks!
-45x^4+21x³+36x
= x (-45x³+21x²+36)
Now you’ve factored out the variable, you can go on with prime factorisation:
= 3x (-15x³+7x²+12)
= 3x [ 3 (4-5x³) +7x² ]
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asks:
6x + 3X -14 = 5
I hope I got what you meant right, that’s how I’d solve it
Remember that you mainly have to focus on how to get the whole equation equals to zero.
asks:
could i have some help with: N2O4 ⇄ 2NO2the equilibrium constant is Kc= 0.2 what are the equilibrium concentrations if one mole of N2O4 is initially placed in a one-liter box? (a simplifying assumption should NOT be used)
Ok, I wrote it all out but keep in mind that if one mole of N2O4 is in a 1 liter box, its concentration would be 1 mol/L, or 1 M.
(I made an error in computing the concentration of NO2 if x had equaled -0.25. [NO2] at equilibrium would have been -0.5. Regardless, this is impossible because you can’t have negative concentrations!)
For those who are not aware, a “simplifying assumption” is assuming that x is minimal in 1-x so that the quadratic is not required to solve (if the assumption had been made, the equation would’ve been simply 0.2=([2x]^2)/1. To see a problem in which an assumption like this is made and how to check if your assumption is correct, see this post.)
Lastly, I know this confused me with ICE problems— remember that since you have 2 moles of NO2, you have to change it by a factor of 2x. Similarly, if you had changed N2O4 by a factor of 0.68, you’d have to change NO2 by 1.36).
-Nic
1. To find the area of a triangle if you know SAS (Side - Angle - Side), you use the formula
You just put the numbers in:
and solve it:
which makes 601.2 mm^2 the correct answer.
2. law of sines:
This means that either two of them are equal, so you just plug in the numbers given (which are A, B, and a) and solve for b:
which gives you 48.4, so that must be the correct answer.
I hope this helped, if you have further questions don’t hesitate to ask
- Kendra
QUESTION: in triangle STU, s=9, t=15, and angle U is 37 degrees. Find u.
-Nic![QUESTION: Find the pH of HCN at equilibrium.
I solved the problem above but here are some extra explanations:
As you know, your dissociation constant, or Ka, is [H][A]/[HA]. You just plug in H, CN, and HCN.
Use an ICE chart to keep track of the change of volumes and then plug your equilibrium values for each into the Ka equation.
Usually, you can assume that your x value is very tiny so you can just pretend it’s (x^2)/2 so you don’t have to use the quadratic. Remember to check to make sure your assumption is correct once you’ve found x. If the number if off by more than 0.5%, you have to use the quadratic with your original equation.
Since [H]=x, you can just take the negative log of your x value and find the pH!
-Nic](http://24.media.tumblr.com/a5a7bf3ecf4b6d799a3e813e056bade5/tumblr_mmx34504h21rhypoio1_1280.jpg)
QUESTION: Find the pH of HCN at equilibrium.
I solved the problem above but here are some extra explanations:
As you know, your dissociation constant, or Ka, is [H][A]/[HA]. You just plug in H, CN, and HCN.
Use an ICE chart to keep track of the change of volumes and then plug your equilibrium values for each into the Ka equation.
Usually, you can assume that your x value is very tiny so you can just pretend it’s (x^2)/2 so you don’t have to use the quadratic. Remember to check to make sure your assumption is correct once you’ve found x. If the number if off by more than 0.5%, you have to use the quadratic with your original equation.
Since [H]=x, you can just take the negative log of your x value and find the pH!
asks:
If I have 78% in an exam worth 40% of a module, what is my current percentage out of 100? For example what do I already have and what do I need in the 60% exam to pass. Please show how you work this out
so it seems like that 78% is the only grade you’ve received so far
here’s how you weight grades
you have two components of your grade, the 40% one and the 60% one. both of these together make up your entire grade
0.4(y) + 0.6(x) = FINAL GRADE
since that 78% is worth 40% of your grade, you can go ahead and multiply it by 0.4
78(0.4)=31.2
and then whatever you get on the final would be multiplied by 0.6
x(0.6)=0.6x
and these two added up would make your final grade
31.2 + 0.6x = FINAL GRADE
so let’s say the passing grade is a 70
31.2 + 0.6x = 70
0.6x = 38.8
x=64.67
and that’s what you’d have to get on the 60% exam to pass
basically just plug in whatever grade you need to pass and then solve for x :)
-Nicasks:
Circle A has a midpoint of (2,-1) and a radius of 5 and circle B has a midpoint of (8,-9) and radius 15. I have to prove that they touch but I keep getting that the length of AB is half that of the combined radii?
Okay let’s see… I drew it out, and that shows that they touch.
You could also use the distance formula, which is 
and that gives you 10 (work shown below). Now if the sum or the difference of the two radii of the circles equal that distance, then the two circles touch.
Radii: 5 and 15.
15 - 5 = 10.
Voila! Your circles touch!
I would hope this makes sense. If you need further clarification, feel free to send in another ask :)
- Kendra
ohhi-school asked maths-help:
asks:
what is the radius of a circle with the area of 728?
To calculate the area of a circle you need the formula A = pi * r2
so r = sqrt(A/pi). Given A = 728, r = 15.2 (approx value).
Hope this helps!
-Alex
The Wonders in π - Jason Latimer
