Firstly, the cube root of 17 is 2.571281591 which is 2.57 to 3.s.f.
Initially, I thought about just approaching this problem using the Newton-Raphson Method when x0=2. In which case; x^3=17
x^3-17=0
Using the Newton-Raphson iterative formula xn+1=xr-f(xn)/f’(xn)
f(x)=x^3-17
f’(x)=3x^2...
Thank you for your reply. Yes, I found found that with nonlinear inequalities. Graphs certainly have been helpful to my understanding this topic better. Although, I still find it a little confusing.
Thank you for your suggestion for the alternative method of solving the second part to this...
Thank you for your reply. For 1, yes sorry my workings should simplify -3x+2-1/x<0 by converting to a fractional form;
-3xx/x+2x/x-1/x<0
-3x^2+2x-1/x<0
Multiply both sides by -1 to reverse the inequality;
3x^2-2x+1/x>0
Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3.
Yes, I...
1. I think the question is asking where is the graph of |3x-2| below the graph of 1/x.
To sketch the graph of y= |3x-2| draw the line of y=3x-2 and reflect the section with negative y-values in the x-axis. Alternatively, I could set 3x-2 ≥0, meaning |3x-2|=3x-2 so draw the line of y=3x-2. Then...
cot^2θ+5cosecθ=4
cot^2θ+5cosecθ-4=0
cosec^2θ+5cosec-4-1=0
cosec^2θ+5cosec-5=0
Let u=cosecθ
u^2+5u-5=0
Solve using the quadratic formula;
u=(-5± 3√5)/2
u=(-5+ 3√5)/2=0.8541...
Substitute cosecθ=u
Therefore, cosecθ=0.8541
1/sinθ=0.8541
sinθ=1/0.8541=1.170... which is not true since sin x cannot be...
Thank you very much for your informative reply. Right, I may add the domain and range of each function for f(x) and g(x) in part a. Really, I used the notation fg(x) and gf(x) because this was what the question specified and additionally what I had been taught in my textbook, which explained...
1. a.
fg(x)=2(1/2(x-1))+1
fg(x)=2(x/2-1/2)+1
fg(x)=x-1+1
fg(x)=x
gf(x)=1/2((2x+1)-1)
gf(x)=1/2(2x+1-1)
gf(x)=x+1/2-1/2
gf(x)=x
The functions functions f(x) and g(x) are inverses of each other. This can be demonstarted by
f(x)=2x+1
y=2x+1
x=2y+1
x-1=2y
(x-1)/2=y
Thus, y=1/2(x-1) = g(x)
And...
Thank you for your reply, sorry I should have used brackets.
Question b:
y=(4x+17)/(x+2)
y=(4x+17)/(x+2)undergoes this first transformation to become y=(4x+17)/(x+2-4)
Transform -4 into a fraction, -4(x+2)/(x+2)=(-4x-8)/(x+2)
y=(4x+17)/(x+2)(-4x-8)/(x+2)
Since the denominators are equal combine...
a. y=x^2 undergoes transformation 1 to become y=(x+2)^2
y=x^2+2 undergoes transformation 2 to become y=3(x+2)^2
y=3(x+2)^2 undergoes transformation 3 to become y=3(x+2)^2+4
So would the equation of the resulting curve be y=3(x+2)^2+4? I am very uncertain when it comes to performing...
I have just looked in the toolbar and have seen what you mean, in addition to the option to insert symbols. Thank you for the suggestion, that is so useful and will definitely allow me to type in a more readable format 👍
Thank you very much for your reply.
So for the first part of the question would I find how much is in the account after 10 years by;
bn =2000*1.06^10
Thus, b10 =2000×1.06^10 = £3581.69539 ~ £3581.70
Then for part 2;
bn=2000*1.06^n
(Replace the inequality with an equation)
Solve bn=5000...
To find how much would be in the account after ten years, let the balance in the account at the start of year n be bn.
Then b1=2000
I believe that this a compound interest problem.
Common ratio r = 1.06
bn =2000*1.06^n−1
Thus, b10 =2000×1.06^9 = £3378.95791
The balance of the account at the...
1. When n=1,
u1+1=3-1/3(u1)
u2=3-1/3(3)
u2=2
When n=2
u2+1=3-1/3(u2)
u3=3-1/3(2)
u3=7/3
When n=3
u3+1=3-1/3(u3)
u4=3-1/3(7/3)
u4=20/9
The common ratio is defiend by r=un+1/un, but this is different between the terms, i.e. u2/u1=2/3 whereas u3/u2=(7/3)/2=7/6
Have I made a mistake?
2. A...
1. The factor theorem states that (x-a) is a factor of f(x) if f(a)=0
Therefore, suppose (x+1) is a factor:
f(-1)=3(-1)^3-4(-1)^2-5(-1)+2
f(-1)=0
So, (x+1) is a factor.
3x^3-4x^2-5x+2=(x+1)(3x^2+...)
Expand the RHS = 3x^3+3x^2
Leaving a remainder of -7x^2-5x+2
3x^3-4x^2-5x+2=(x+1)(3x^2-7x+...)...
Thank you for reply. Admittedly, I am still a little confused but far less than I was. I think I understand what you mean though.
So I should use the binomial formula to expand (1+1/3)^18=1+6x+17x^2+272x^3/9+340x64/9...
Then multiply this by the first bracket:
(3-5x)(1+6x+17x^2+272x^3/9+...)...
Sorry I am supposed to be converting from 0.2 mol dm^-3 to g cm^-3 (the reciprocal quantities) so would I still divide by 1000?
54 g dm^-3 = 0.054 g cm^-3 ?
Thank you for your reply. Sorry yes, I see my mistake I think.
If there are 1000cm^3 in 1 dm^3 then there will be a greater volume of solute in 1cm^3 than 1dm^3?
So could the concentration be found by;
1 g dm^3=270 mol dm^3 /5=54g dm^3
Then divide this result by 1000 to convert dm^3 to cm^3:
54...
Thank you very much for your reply and for your advice, I admit I will probably use your statement. It is a question my physics tutor sent to me in a quantum physics revision package of questions, but I had not come across a homogenous problem to this before, at least not within the context. If...
Thank you for your reply and your continued insight.
2. c) Sorry my mistake, clearly I misused intensity here.
Question 3) "example" was the wording from the original question. Oh right, I will correct my average value also, thank you for remarking on that.
Can I verify that it would be...
Thank you very much for your reply and for pursuing this further. My apologies but I am a little confused what the correct angle to use would be when finding the intensity?
Thank you for your reply. No I suppose not, I am just uncertain how to calculate this. I multiplied by 1000 since I know that the to convert between dm^-3 and cm^-3 you would do so. Would I actually only multiply by 10, since this is how to convert from dm to cm?