help please (1 Viewer)

[Yoko Hono]

hi having a few problems with mathematics revision

1. solve the following equations where (0<x<2pi)

(a) cos x = 1/2

(b) tan x = -1

2. Find the area enclosed by the curve y = cos2x, the x axis, and the lines x = - pi/6 and x = +pi/6

3. A pile of sand dumped at the start of a long road has to be distributed in truckloads at 800metre intervals along the road.
If there are 15 truckloads of sand in the pile, calculate how far in total the truck has to travel in order to deliver the sand.

4. Calculate the compound interest on $50,000 compounding every 6 months at 7%per annum for 2.5 years.

5. Tom's father paid $500 into an account on the day Tom was born after that he paid $500 into the account on Tom's birthday until Tom's 18th birthday. If the account accrued interest at 8% p.a.
compounded monthly calculate how much Tom would receive on his 18th birthday.

6. Find the exact value of 3 2+5x dx
S x^3
1


7. Find all values of radiants for 0<radiants<2pi
(i) 2cosradiants=sq.root 3
(ii) 2sin(radiants-pi/4)
(iii) sec squared(radiant/2) = 2
(iv) sin2radiant=cos 2 radiant

8. A hospital patient receives 10mg. dose of medicine at the beginning of each day. Prior to the next dose the amount of medicine in the patients body reduces to 3/5 of the amount present find

(i) the amount of medicine in the body on the 7th day just after the next dose is given.

(ii) does the medicine present in the body ever exceed 25mg.

 

[Kulazzi]

for 1 (a) and (b) press shift cos/tan and then the number 1|2/-1

for 2 and 3 I have no idea/can't be bothered

For 4, use the A = p(1+r/100)^n and sub in each one into the formula

5 - I think same as four but b/c it's compunded monthly for 18 years, n = 18/12

6. Isn't that the same q you asked in your other thread about the 3 changing into the 2?If it is then go have a look there again

7. No idea

8. No idea

 

[acmilan]

Actually for 1 (a) and (b) there would be more than one solution for each (assuming by 0<2pi you actually mean 0<x<2pi)

 

[Trev]

hi having a few problems with mathematics revision

1. solve the following equations where (0<x<2pi)

(a) cos x = 1/2

(b) tan x = -1

(a) basic angle x = 60 degrees, and 1/2 is positive therefore lies in quadrants 1 and 4 (CAST). So x = 60, 300 (degrees).

(b) basic angle x = 45 degrees, and -1 is negative therefore lies in quadrants 2 and 4 (CAST). So x = 135, 315 (degrees).

 

[Trev]

2. Find the area enclosed by the curve y = cos2x, the x axis, and the lines x = - pi/6 and x = +pi/6

[S - integral sign]
-π/6
S cos2x.dx
π/6

-π/6
2 S cos2x.dx
0

2[1/2sin2x] π/6 to 0

sin(2*π/6) - sin(2*0)

= (sqrt3)/2 units²

 

[Trev]

3. A pile of sand dumped at the start of a long road has to be distributed in truckloads at 800metre intervals along the road.
If there are 15 truckloads of sand in the pile, calculate how far in total the truck has to travel in order to deliver the sand.

Not too sure what this is asking, but this is what I would do.
Let one truckload of sand = x
since 15 truckloads, 15x, there would be 14 intervals.
So distance truck has to travel is 15x + 14(800).

 

[Trev]

4. Calculate the compound interest on $50,000 compounding every 6 months at 7%per annum for 2.5 years.

= 50000[1 + (7/2)/100]^(2.5*2)
Tell me if you do not understand what I just did, you should be able to follow it through.

 

[Trev]

5. Tom's father paid $500 into an account on the day Tom was born after that he paid $500 into the account on Tom's birthday until Tom's 18th birthday. If the account accrued interest at 8% p.a.
compounded monthly calculate how much Tom would receive on his 18th birthday.

'up until Tom's 18th birthday', I assume this means not including, therefore it would be 18 times he would deposit $500, if it means including his 18th birthday it would be 19 times. So;

A1 = 500[1 + (8/12)/100]^12
A2 = 500[1 + (8/12)/100]^24 + 500[1 + (8/12)/100]^12
A3 = 500[1 + (8/12)/100]^36 + 500[1 + (8/12)/100]^24 + 500[(8/12)/100]^12
continuing till 18th deposit:
A18 = 500[1 + (8/12)/100]^216 + 500[1 + (8/12)/100]^204 + ... + 500[(8/12)/100]^12
= 500{[(8/12)/100]^12 + ... + [1 + (8/12)/100]^204 + [(8/12)/100]^216]}

With geometric series
a = [(8/12)/100]^12
r = [(8/12)/100]^12
n = 18
You can do the rest using formula.
a[(r^n - 1)]/(r - 1)
I'm actually not too sure it is right, so correct me if I am wrong - i'm a bit rusty on this stuff.

 

[Trev]

6. Find the exact value of 3 2+5x dx
S x^3
1

[S = integral sign]

3
S (2+5x)/(x³) dx
1

3
S 2x^-3 + 5^-2 dx
1

[ -x^-2 - x^-1 ] from 3 to 1

and you should be able to do the rest, if not tell me.

 

[Trev]

8. A hospital patient receives 10mg. dose of medicine at the beginning of each day. Prior to the next dose the amount of medicine in the patients body reduces to 3/5 of the amount present find

(i) the amount of medicine in the body on the 7th day just after the next dose is given.

(ii) does the medicine present in the body ever exceed 25mg.

(i)
A1 = 10
A2 = 10*3/5 + 10
A3 = 10*(3/5)² + 10*(3/5) + 10
Therefore;
A7 = 10*(3/5)⁶ + 10*(3/5)⁵ + ... + 10(3/5) + 10
= 10[(3/5)⁶ + (3/5)⁵ + ... + 3/5 + 1]
Geometric series
10 {1[(3/5)^7 - 1]/(3/5 - 1)}
= 24.30016 mg of medicine in body on 7th day.

(ii) Sum of infinite series a/(r-1)
Gives
= 10[1/(1 - 3/5)]
= 25, so it will never exceed 25mg.