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...
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...
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...
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...
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+...)...
a) Evaporation will remove water from the test tubes as it turns into water vapour, meaning that the solution will have a greater solute concentration and thus an increased osmotic potential which results in a more negative osmotic potential. Consequently this lowers the solution's water...
1. The 4th line from the left, being the aqua blue line, corresponds to a wavelength of 486 nm, as blue light has a wavelength in the range 450-495 nm.
2. This is where I am having the most difficulty, I have tried to answer the question comprehensively but I am not satisfied with my answer.
In...
1. a)I have plotted the graph on desmos and attached an image here.
b i. The threshold frequency is equal to the x-intercept ~ 5.6*10^14 Hz
ii. The work function is equal to the y-intercept ~ -3.75*10^19 J (would it be correct to state that this value is negative?)
c. Convert to eV;
3.75*10^19...
1. I believe that the gravitational field strength would decrease because it is inversely proportional to the square of the distance from the centre of the Earth, g∝1/r^2.
Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the...
1. The centripetal force is equal to F= mv^2/r.
The velocity of the earth can be found by:
V=2πr/T
T=1 day = 24 hr*60min*60sec=86400 s
v=2π*6.4 x 10^6/86400 s
v=465.4211 ... ~465 ms^-1 to 3.s.f
Therefore, F=1*465/6.4 x 10^6
F=98/1280000=7.265626 *10^-5 ~7.3 *10^-5 N
Would this be correct since...
a. V=-GM/r
V=-6.67*10^-11*6.0 x 10^24/6.4 x 10^6
V grav = -62531250 ~ -62.5M Jkg^-1
b. To find the gravitational potential 200 km above the surface of the Earth;
r=6.4 x 10^6 +2*10^5 m=6.6*10^6
V grav=-6.67*10^-11*6.0 x 10^24/6.6*10^6
V grav= -60636363 ~ -60.6 M Jkg^-1
Can I check that it is...
1. The satellite would be in a jovian-synchronous orbit,
Rearranging the formula for the orbital period in terms of r, since T^2 is proportional to r^3:
T^2=4π^2r^3/GM which becomes r^3=(GM/4π^2) T^2
M(mass of Jupiter)=1.89 x 10^27
G=6.67*10^-11 m^3kg^-1s^-2
T=9 hours and 55 minutes =...
1. Newton's Second Law states F=ma and the formula for centripetal acceleration is v^2/r
Therefore, F= mv^2/r
Would this be complete, I just feel that I should need to do something further but I am not sure what?
2.F=mv^2/r
Gravitational force = GMm/r^2
Gravity is the cause of centripetal...
1. Since the gravitaional field strength is 1/6 of that on Earth:
W=mg
W=90*9.81/6
W=90*1.635
W=147.15 ~ 147 N
2. ∆Ep=mg∆h
∆Ep=90*1.635*50
∆Ep=7357.5 J
I do not now whether this method would be suitable and if I should have instead used the formula for gravitaional Potential, V grav=-Gm/r?
3...
So I think I may be overcomplicating this problem but I realise that in order to find the x^3 term it will be the product of the two binomials, ie. x^1*x^2=x^3. The coefficient of x^3 will be the coefficient of x^1 in the first bracket multiplied by the coefficient of x^2 in the second bracket...
a. I have just plotted the graph using desmos and attached an image here. Clearly, there are two values of x that satisfy the equation in the range. Do I need to add anything to this statement, I feel the response is a little brief for the question?
b. Using the trigonometric identities;
tan...
Question 1;
a. sin θ=√3/2
θ=arcsin √3/2
θ=π/3 rad
sin √3/2=60 degrees
60 degrees *π/180=π/3 rad.
To find the other solutions in the range, sin θ=sin(π-θ)
π-π/3=2π/3
The solutions are π/3 and 2π/3 in the range 0 ≤θ ≤2 π
b. cos2θ=0.5
2θ=arccos 0.5
2θ=π/3 rad
Divide both sides by 2;
θ=π/6 rad...
1. Using the formula for the arc length; s= θr
I have endeavoured to find the angle AOB sine both the arc length and radius are known;
11= θ*8
θ=11/8=1.375 rad
I actually do not think that this can be correct as it seem to simplistic a response. Have I misinterpreted the question or used the...
Question 1;
a. sin θ=√3/2
θ=arcsin √3/2
θ=π/3 rad
sin √3/2=60 degrees
60 degrees *π/180=π/3 rad.
To find the other solutions in the range, sin θ=sin(π-θ)
π-π/3=2π/3
The solutions are π/3 and 2π/3 in the range 0 ≤θ ≤2 π
b. cos2θ=0.5
2θ=arccos 0.5
2θ=π/3 rad
Divide both sides by 2;
θ=π/6 rad...
Question 1; Method 1
If the sum of the first four terms is 139 then S4=139
139=1/2(4)(2a+(4-1)d)
139=2(2a+3d)
139=4a+6d----- [1]
The part of this question that is confusing is the "the sum of the next four terms is 115".
Would this mean that S8=S4+115=139+115=254?
In which case...
i. Using Newton's 2nd Law, F = m a
consider the motion of the entire system, so the car, caravan and towbar an be thought of as a single object.
The tension can ignored as it is an internal force.
Braking fore + resistive forces = mass * acceleration
Braking force + 200N +150 N=(1000+1500)*(0.5)...
So I have attempted to plot the scatter diagram. My first query is does the question intend for you to include both subsets of data on one axis, (which I have plotted on the x-axis) or rather does it demand two separate diagrams to investigate if there is any correlation, or a single diagram? I...
Question 1:
a) T' is the complementary event of T
Therefore, T'=1-T
In set T = {3,6,9,12}
P(T)=4/12 =1/3
P(T')=1-1/3=2/3
b) The addition rule states; P(A ∪ B)=P(A)+P(B)-P(A⋂B)
Therefore, P(S ∪ E) = P(S)+P(E)-P(S⋂E)
S={1,4,9}
P(S)=3/12=1/4
E={2,4,6,8,10,12}
P(E)=6/12=1/2
(S⋂E)={4}
P(S⋂E)=1/12...
I have attached a photograph of my workings. I do not know if I have arrived at the right solution, nor whether this is the gradient of f(x) at point P.
I think I seem to overcomplicate these problems when thinking about them which makes me lose confidence in my answers. Thank you to anyone who...
The displacement of the particle is;
s= ∫v dt
s= ∫4-3t^2 dt
s=4t-t^3+c
When the particle is at the Origin, t=0;
0=4(0)-(0)^3+c
c=0
So this becomes;
s=4t-t^3
The particle next passes through the origin when;
4t-t^3=0
Factor out the common term -t;
-t(t^2-4)=0
Rewrite t^2-4 as t^2-2^2 and factor...
Equation 1:
Where t1=time spent on motorway
Where t2=time spent on country roads
t1+t2=4
Equation 2:
Using distance = speed * time
200 = 70*t1+40*t2
Rearrange equation 1 in terms of t1;
t1=4-t2
Substitute the rearranged form of equation 1 into equation 2:
200=70(4-t2)+40t2
200=280-70t2+40t2...
Well, I understand that according to the conservation of momentum the total momentum of a system is conserved for objects in an isolated system, that is the sum of total momenta before the collison is equal to the sum of momenta after the collision.
In this case, the momentum of the object...