Can you solve this (2) Mathematics questions in the pictures below?

Solve it correctly and you maybe rewarded.


Cc: Lalasticlala

So for ur mind u wise

Post ur assignment on NL looking for free answers ba

BTW wat is the winner goin to get

Sirsoftness:
So for ur mind u wise

Post ur assignment on NL looking for free answers ba

BTW wat is the winner goin to get

Not my assignment! I don pass all this level.

I have kids who are in schools. This is a recent exams questions in the Uk that 85% of the students failed.

Reward is #5000 worth of airtime of your choice.

Its hard sha
For my level





Wish I payed attention in Math class

MosakuAW:


Not my assignment! I don pass all this level.

I have kids who are in schools. This is a recent exams questions in the Uk that 85% of the students failed.

Reward is #5000 worth of airtime of your choice.

I'm holding you to your words, even if you don't deliver, maybe someone might learn a thing or two.

I've solved the second question.

t2 = 22.75.
Using the recurrence formula.
tn+1= (0.75 * tn )+ 13.
Where tn = 13.
Substituting tn in the recurrence formula we get:
t1+1 = (0.75 * 13)+ 13
t2 = 9.75 +13
t2 = 22.5

For the second part of the question, the toad will get out in exactly 12 days after travelling 50.35 metres (in 2dp)
I did it by hand via iteration.
There's a summation formula for recurrence but I'm too high right now.

The frog will remain down there because the recurrence
formula for the frog converges very slowly.



A better method to answer the second part of the question is to find the limits of the two recursive formulas and see if any surpasses 50.


For f_n+1 = f_n / 3 + 32
To evaluate this limit, we will do shorthand induction to first show that the sequence a_n is increasing and bounded above.

First we note that a₁ = 32, a₂ = 42.56, a₃ =46.04

In each case, our sequence appears to be increasing, and we will prove this with mathematical induction. Our base case a₁ < a₂ is already shown above. Suppose that a_k < a_k+1 We want to show that a_k+1< a_k+2 We get that a_k+2 = 1/3 a_k+1 + 32 > 1/3 * a_k = a_k+1 and so a_n is increasing.

Now we need to show that a_/n is bounded. We assume that a_n is bounded above by 96 since a₁ = 32 < 96. Now suppose that a_k < 96 We want to show that a_k+1 < 96
So we have that a_k+1 = a_k/3 + 32 < (1/3 * 96) + 32 =64. Therefore a_n < 96.

Now recall that since f(x) = x/3 + 32 is a continuous function, we have that:
L, which is the lim_x->oo of a_n+1 = lim_x-2>oo of (a_n /3 + 32)
L = L/3 + 32
L = 48.
Therefore since the limit of the function of the frog is less than 50, even if it moves for the next 10 billion years it'll still be less than 50m.

For the toad:
L = 3L/4 + 13
L = 52.
The limit of the toad is 52which is greater than 50, so the toad scales through the hole.

Now that I am sober.

On to question 1.

I'll start from the (b) part, as that's the part that I understand intuitively.
To find the minimum possible time, an the corresponding x value.
I'll also try to explain as best as I can.
The minimum point is one of the turning points of a function, other turning points are the maximum point and inflexion point.
Consider a function f(x)
The turning points of a function lie at f'(x) which is the derivative of y with respect to x, or in Leibniz notation dy/dx.
At the turning point, dy/dx = 0.

Analytically:
If f'(a)=0 and f"(a) is positive then there is a minimum value at x=a
If f'(b)=0 and f"(b) is negative then there is a maximum value at x=b.
Where f'(x) and f"(x) are the first and second derivatives of f(x) respectively. and a and b are the roots of the equation f"(x)=0.

F(x) or T(x) in this case= 5√(36 +x2) + 4(20-x)
dy/dx = 5(36+x^2) +4(20-x)
=5*u1/2 + 80 -4x, where u= 36+x2
From the chain rule, dy/dx =dy/du * dy/dx
= 5 * 1/2√u * ( 2x) -4
=5* 2x/2√(36+×2) -4
=5x/√(36+×2 - 4

And that is f'(x)
Will continue in next post.

Continuing from above...
F'(x) = 5x / √(36+x²) -4
Equating it to zero because we want to find a and b, the roots of the equation. This equation is linear so there'll only be one answer.
5x/√(36+x^2) -4 =0
5×/√(36+×^2) = 4
Cross multiplying.
5× = 4* √(x^2 +36)
Squaring both sides to remove the square root.
(5x)² = [4*√(x^2+36)]²
25x² = 16 * (x^2 +36) (The square cancels the square root)
Expanding the brackets.
25x² = 16x² +576
Collecting like terms
25x^2 - 16×^2 = 576
9x² = 576
Dividing both sides by 9.
x² =64
Taking square root of both sides.
x = 8.
As expected there is only one root, we'll call it a.

Moving on..

Recall we have two conditions to satisfy for the minimum point of a function.
We have f'(a) =0 which we just showed in the last post. You can confirm by substituting 8 for x in f(x), you'll surely get 0.
The second condition: f"(a) has to be positive.

Also recall that f"(x) is the second derivative of a function or d²y/dx² in Leibniz notation.

I'll skip the steps for finding the second derivative of f(x), basically you just differentiate f'(x)
If you do that you'll get:
f"(x) = 180/ (x² +36)^3/2

Now we substitute x =8 into this , if we get a positive answer then there is a minimum at x =a.
If you substitute x=8 into this you get 9/50, which is positive.

Now the two conditions are satisfied, so we do indeed have a minimum value of f(x) at x = 8.
To find the minimum value of f(x) you have to substitute x=8 into the original equation.
f(x) = 5√(36+×^2) +4(20-x)
f( 8 ) = 5√(36 +8²) + 4(20 - 8 )
f( 8 ) = 5√(36 +64) +4(12)
f(8 ) = 5√(100) +48
f( 8 ) = (5 * 10) +48. (The square root of 100is 10)
f( 8 ) = 50 + 48.
f( 8 ) = 98.

Therefore the minimum value of f(x) is 98.
Note that in the question the unit of T(x) is deciseconds(tenth of a second)
So our minimum value will be . 9.8 seconds at x=8.
That's all for the (b) part.
We move to the simpler (q) part.

The (a) part is simple, all we need to do is consider the diagram.
(I) If it does not travel on land then that means it travels in the water.
Also notice that the animal is 20 metres away.
According to the diagram the crocodile is at PX.
So the distance it'll take it to reach px is 20
Substitute 20 for x in the equation and you get
T(20) = 5√(36 + 20²) + 4(20-20)
T(20) = 5√(36+400) +4(0)
T(20) = 5√(436)
T(20) =5 * 20.88
T(20) = 104.4 deciseconds if it follows the water.
This is the same as 10.04 seconds.


(ii) From the question the shortest distance is at Px.
And from the diagram the crocodile is at PX.
So it that means x =0. Since it is already on land.
Now substitute x=0 into the original equation
T(0)= 5√(36 + 0²) + 4(20 -0)
T(0) = 5√36 + 4(20)
T(0) = (5 * 6) + 80
T(0) = 30 + 80
T(0) = 110 deciseconds.
Which is the same as 11 seconds.

That's all.
OP, I'm done.

Now a moment of silence for the poor Zebra.

This is a direct download link to the paper if anyone is interested.
The questions are actually very simple.