Mathematics Marathon HSC 09 (2 Viewers)

[s2Vicki]

Question:

Find the maximum product of 2 numbers whose sum is 25.

 

[00iCon]

question:

Find the maximum product of 2 numbers whose sum is 25.

12.5^2
=156.25?

 

[00iCon]

if a,b,c are roots of x^3 - 4x^2 +1=0
find:
i) a+b+c
ii)1/a+1/b+1/c

 

[xFusion]

ummm. it cant?
x=ln y
rite guys?

My bad. I got that straight out of a random page of a textbook. Dumb book with misprints =.=

 

[xFusion]

if a,b,c are roots of x^3 - 4x^2 +1=0
find:
i) a+b+c
ii)1/a+1/b+1/c

i)-b/a =4
ii)c/a=ab +ac+bc =0
1/a+1/b+1/c =0/4 =0
Question: f(x)=√(1 +x^2) where -1<=x<=4
Using 6 function values, using Simpsons rule, find the approximate area under the curve.

 

[00iCon]

i)-b/a =4
ii)c/a=ab +ac+bc =0
1/a+1/b+1/c =0/4 =0
Question: f(x)=√(1 +x^2) where -1<=x<=4
Using 6 function values, using Simpsons rule, find the approximate area under the curve.

Correct!

I refuse to use simpson's rule!

 

[AlexJB]

Question:

Find the maximum product of 2 numbers whose sum is 25.

x+y = 25
y = 25-x
f(x) = xy = x(25-x) = 25x - x^2
f'(x) = 25 - 2x
when f'(x) = 0, x=12.5
f''(12.5) = -2, therefore max.
y = 25-12.5 = 12.5
hence x = 12.5, y=12.5

if a,b,c are roots of x^3 - 4x^2 +1=0
find:
i) a+b+c
ii)1/a+1/b+1/c

i. a + b + c = 4
ii. 1/a + 1/b + 1/c
= (ab + ac + bc) / abc
= 0

Question:

A radioactive isotope has a mass of 80g. It decays to a mass of 50g in 10 hours. After t hours its mass M grams is given by M = 80e^(kt) for some constant k < 0.

i. Find the value of the decay rate k to 3 decimal places.
ii. Find the rate of decay of the isotope after 4 hours, correct to 1 decimal place.
iii. Find the half-life of the isotope to the nearest minute.

 

[s2Vicki]

i)-b/a =4
ii)c/a=ab +ac+bc =0
1/a+1/b+1/c =0/4 =0
Question: f(x)=√(1 +x^2) where -1<=x<=4
Using 6 function values, using Simpsons rule, find the approximate area under the curve.

Answer:
table of values:
x -1 0 1 2 3 4
f(x) √2 √1 √2 √5 √10 √17

√[1+(x^2)] dx = (1/3)[(√2+√17)+2(√2+√10)+4(√1+√5)]
= 9.2115...

Question: 2(x^2) + pq + q = 0 has one root 3 times the other. Show that 3(p^2) = 32q

 

[xFusion]

i)when t=10, m=50
k=-0.047
ii)dM/dt=k.80e^(kt)
when t=4, dM/dt=-3.116 ---> rate of decay =3.1 g/hr
iii)half-life when m=40
=14hrs 45min
Question: series 2,4,8,16,32.... find the 20th term in the series

 

[s2Vicki]

i)when t=10, m=50
k=-0.047
ii)dM/dt=k.80e^(kt)
when t=4, dM/dt=-3.116 ---> rate of decay =3.1 g/hr
iii)half-life when m=40
=14hrs 45min
Question: series 2,4,8,16,32.... find the 20th term in the series

a=2, r=2

Tn = ar^(n-1)
T20 = 2*(2^19)
= 1 048 576

 

[xFusion]

Answer:

Question: 2(x^2) + pq + q = 0 has one root 3 times the other. Show that 3(p^2) = 32q

Are you sure that works out? because i got both roots as zero and then couldn't find q.

 

[addikaye03]

New Question:

Find the sum to n terms of the series

a) 1+5+9+...

b) 3+7+11+...

Hence or otherwise, find the sum of the series

1-3+5-7+9-11+.... to i)100 terms ii)101 terms

 

[s2Vicki]

Are you sure that works out? because i got both roots as zero and then couldn't find q.

it should...

 

[xFusion]

it should...

Can you show me how to do it?
cause i got let roots = @, 3@
-b/a=4@=0
c/a=(pq +q)/2 =3@^2
q(p+1)=0, p=-1
and thats it lol

 

[s2Vicki]

New Question:

Find the sum to n terms of the series

a) 1+5+9+...

b) 3+7+11+...

Hence or otherwise, find the sum of the series

1-3+5-7+9-11+.... to i)100 terms ii)101 terms

Answer:
a) a=1, d=4
Sn = (n/2)[2a+(n-1)d]
= (n/2) [2(1)+(n-1)4)
= (n/2) [3+4n-4)
= (n/2) [4n-1]
= n[8n-2]
= 8(n^2)-2n

b) a=3, d=4
Sn = (n/2) [2(3)+(n-1)4]
= (n/2) [6+4n-4]
= (n/2) [4n+2]
= n(8n+4)
= 8(n^2)+4n

i) a=-4, d=-8
Sn = (100/2)[2(-4)+(100-1)(-8)]
S100 = -40 000

ii) -40 000 -8 = -40 008

 

[s2Vicki]

Can you show me how to do it?
cause i got let roots = @, 3@
-b/a=4@=0
c/a=(pq +q)/2 =3@^2
q(p+1)=0, p=-1
and thats it lol

lol i don't know how to do it. that's why i asked.

 

[AlexJB]

it should...

You wrote the question wrong.

Proof:
Let b = 3a (i.e. 2nd root 3 times the first).

Your equation: 2(x^2) + pq + q
a + 3a = 0 (no x^1)
4a = 0
Therefore a=0, 3a = 0, two roots are 0, not true.

Invalid question =p

----

edit: p & q are constants or not? If not, then p(q + 1) = 0
p = 0 or q = -1

----

edit2: misread question. let me actually look at it.

---

edit3: im pretty sure there's nothing you can do with that.

 

[xFusion]

New Question:

Find the sum to n terms of the series

a) 1+5+9+...

b) 3+7+11+...

Hence or otherwise, find the sum of the series

1-3+5-7+9-11+.... to i)100 terms ii)101 terms

a)n/2*(2 +4(n-1))
b)n/2 *(6 +4(n-1))
i)50/2(2 +4(49)) -50/2(6 +4(50-1))
=-100
ii)101

Question: Water flows from a full tank at the rate given by dh/dt= -1.5 sqrt(t)
where h is the depth of the water in meters at any time t minutes.
a)find the equation for h in terms of t
b)how long does it take for the tank to empty?

 

[addikaye03]

a)n/2*(2 +4(n-1))
b)n/2 *(6 +4(n-1))
i)50/2(2 +4(49)) -50/2(6 +4(50-1))
=-100
ii)101

[/b]

correct

ii) -40 000 -8 = -40 008

That's not correct, try counting the in two terms succession and you might see a pattern, the Q is really simple if you see it

 

[s2Vicki]

[/quote]i) a=-4, d=-8
Sn = (100/2)[2(-4)+(100-1)(-8)]
S100 = -40 000

ii) -40 000 -8 = -40 008
And the rest? lol[/quote]

i couldn't be bothered then. but i've done it now. check the post again. i edited it.