Maths problems (1 Viewer)

[Real Madrid]

Show working:

(the attachment)

 

[Real Madrid]

Maths Assistance

See the attachment...

can someone solve it showing all working out?

i got stuck

 

[xclo]

Re: Maths Assistance

h cot 42 = OA
h cot 21 = OB

OB^2 - OA^2 = AB^2

(h cot 21 )^2 - (h cot 42 )^2 = 400^2

result then follows

 

[Aerath]

Re: Maths Assistance

Yup, expand, then factorise the h^2 out, then divide and then square root both sides.

 

[Real Madrid]

Solve : cos^2x - cos 2x = 0 for 0≤x≤2(pi)

The (pi) bit is the symbol for pi.

 

[Real Madrid]

maths problem

Solve : cos^2x - cos 2x = 0 for 0≤x≤2(pi)

The (pi) bit is the symbol for pi.

 

[Just.Snaz]

Re: maths problem

(cosx)^2 - cos 2x = 0
(cosx)^2 - [2(cosx)^2 - 1] = 0, using the double the angle rule
(cosx)^2 - 2(cosx)^2 +1 = 0
(cosx)^2 = 1
(cosx) = +-1
x = 0, pi, 2pi

 

[mikierceg]

Re: maths problem

oh well...
i would help you but going sleep

 

[Real Madrid]

more maths problem

my class is whizzing through 3 unit content and bleh bleh they want to start moving quickly so they can do revision early...so...my teachers a bitch and i have barely any idea to do any of these questions except with the use of year 11 knowledge on derivatives. i practically need help with q2-7

Look at attached

 

[Just.Snaz]

Re: more maths problem

my class is whizzing through 3 unit content and bleh bleh they want to start moving quickly so they can do revision early...so...my teachers a bitch and i have barely any idea to do any of these questions except with the use of year 11 knowledge on derivatives. i practically need help with q2-7

Look at attached

a) monotonic decreasing meaning your gradient, ie, f'(x) < 0
b) for stationary points you make f'(x) = 0

3. f(0) = 1 therefore y-intercept at y = 1
f'(x) > 0 for 0<= x <= 3
meaning the gradient of the curve is positive between x = 0 and x = 3

f'(3) = 0, well you know that stationary points occur at f'(x) = 0 so here, there's a stationary point at x = 3 and they've given you that when you sub x = 3 into f(x) you get f(x) = 5, ie stationary point at (3 , 5)

f'(x) < 0 for values greater than x = 3
and f(4) = 0, therfore at x =4, y = 0, ie, an x-intercept at x = 4

doing these questions should help you do the rest. the essential is
stationary points at f'(x) = 0
inflexion points at f''(x) = 0
x -intercepts at y = 0 [y is f(x)]
y-intercepts at x = 0
f'(x) is the gradient.


for 6, i forgot how to do it.
for 7, you need discriminant.. which is b^2 - 4ac
different real roots means discriminant > 0
real roots means discriminant >= 0
unreal roots means discriminant < 0
one real root or equal roots means discriminant = 0


hope I'm not doing your hw for you..

 

[Real Madrid]

Re: more maths problem

nah you're not its just that i dont know how to approach these problems properly. ty anyway

 

[-tal-]

Re: more maths problem

That's... a lot

2a) No can do
2b) For stationary pts, f'(x) = 0

f'(x) = x^2 - 4 = 0

x = 2 or -2

3) Can't really show you a pic...

4) For turning points, y' = 0

y' = 2ax + b = 0

let y' = f'(x)
f'(3) = 0
2a(3) +b = 0
6a + b = 0 ... 1

f(3) = -11 (point 3, -11 lies on the curve)
f(3) = a(3)^2 + 3b + 7 = -11
9a + 3b = -18
3a + b = -6 ...2


1 - 2
3a = 6
a = 2

Sub a = 2 into 1
b = -12

5) For stat pts, y' = 0

y' = 4x^3 - 16x = 0
x = 2 or -2 or 0

y'' = 12x^2 - 16
x=2, y'' = 48 > 0, therefore it's a min turning point
x= -2, y'' = 48 > 0, therefore it's a min tp
x = 0, y'' = -16 < 0 , therefore max tp

x = 2, y = 0
x = -2 , y = 0
x=0, y = 16

are turning pts

 

[-tal-]

Re: more maths problem

6) wth?

7) let roots be m and v

since m=v, roots are m and m

m + m = (k+3)/1
2m = k + 3
m = (k+3)/2 ..1

m^2 = 4...2

Sub 1 into 2

eh you do the rest im tired and cbf typing up the rest

 

[Real Madrid]

Biology Task Guide

I need to find 2 examples of australian ectotherms and endotherms. I have to describe thier adaptations to assist them with body temperature regulation. Heres the problem, I do not know of any good sites that have such information. Wiki is too irrelevent.

I was thinking of Red kangaroo and hopping mice for endotherms and goannas and bogong moths for ectotherms...

 

[Timothy.Siu]

Re: Biology Task Guide

yeah do them then, just google them and u shud get info

 

[aznvdz]

Re: Biology Task Guide

Your assignment is similar to mine, haha.

 

[Real Madrid]

maths problem

Find all values for x for which the curve f(x)= x^3-3x+4 is decreasing.

When derived f'(x)=3x^2-3

For decreasing curves f'(x)<0

3x^2-3<0
3(x^2-1)<0
x^2-1<0
(x-1)(x+1)<0


now heres where im confused. 1 and -1 are turning points. Now do i test for decreasing curves...?

 

[Aerath]

Re: maths problem

I double dash it.
f''(x) = 6x
sign f''(x) = x
Draw y = x and you'll get a line passing through 0, positive gradient.
That means x<0, concave down, x>0 concave up.

Therefore, x<0 for negativity concavity.

 

[youngminii]

Re: maths problem

Therefore, x<0 for negativity concavity.

We're not finding Concavity, we're finding Decreasing parts of the curve, major difference

Okay well this is the calculus method.. You just test for decreasing ==;;
Sub in 0 (since it's between -1 and 1) into the derivative
3x^2 - 3
0 - 3
-3 < 0
So it's decreasing between -1 and 1
ie. Decreasing when -1 < x < 1

Non calculus method
You should know by now how a cubic looks like
If the coefficient of the largest power is positive (ie this equation)
The curve will start off increasing, then it'll turn and decrease, and turn and increase once again
So, if -1 and 1 are the turning points
Then the curve will clearly be decreasing between those points

Hope that helps

 

[Aerath]

Re: maths problem

Ooooo, yeah, my bad - interpreted the question incorrectly.