All Answers 358108 Answers

This is a sideways parabola

$$\\3y^2-4x=0\\\\
y^2=\frac{4x}{3}\\\\
y^2=4 \times \frac{1}{3}\times x\\\\$$

Focal length = 1/3

focus (1/3, 0)

Laus r****m is 4*1/3 long = 4/3

The ends of it will be (1/3, 2/3) and (1/3, -2/3)

You could also take away 2n from n.

it was 88 over 28 unknown but your logic is correct.

Have a look at this page.

I guess not

6*1000=6000

Have a look at this.     

you could multiply or divide it by negative 1

$$\\\frac{80.6}{4.096\times10^{-12}}\\\\
=\frac{80.6\times 10^{12}}{4.096}\\\\
=19.678\times 10^{12}\\\\
=1.9678\times 10^{13}\\\\$$

Correct to 5 significant figures.

You should 'study' this post.     

First do 19-14 and then add a minus sign to the start of the answer. The number you are left with is b.

The answer is 0.

$$1/4=0.25\Rightarrow2+0.25=2.25$$

1)

$$\\F(x)=x^2+2\\\\
F(x^2)=(x^2)^2+2=x^4+2\\\\$$

3)

$$G(x)=3x^2-2x-1

G(a+b)=3(a+b)^2-2(a+b)-1\\\\
G(a+b)=3(a+b)^2-2(a+b)-1\\\\$$

You can expand and simplify this one.   

$$\\50+218x^3-28+12x^6\\
=50-28+218x^3+12x^6\\
=22+218x^3+12x^6$$

Look here for a general all inclusive answer.

"The way I have drawn my v-t graph from the origin to  point P  is concave up.

If this is correct shouldn't the acceleration (between the origin and P) always be positive and decreasing. "

Yes, you're right, it should.

Thanks you anonymous.  The answerers here all try their hardest to help.  

It is always so nice when people voice their appreciation.

Why don't you join up.  We love having such polite and grateful members and it would be nice to get to know you.      

Alan, I am a little confused.

The way I have drawn my v-t graph from the origin to  point P  is concave up.

If this is correct shouldn't the acceleration (between the origin and P) always be positive and decreasing.  

Thanks very much Alan,  

*** points P and R in the first picture ( x - t ) is a turning point but how did you know that p is the maimum velocity and R is minimum ? ***

I have assumed that P and R are points of inflection.

If this is the case then at P on the x-t graph has the steepest positive gradient hence it will have maximum velocity.

And

R on the x-t graph has the steepest negative gradient hence it will have the greatest negative velocity.

-------------------------------------------------

Remember 3*15,

The gradient on the x-t curve IS dy/dx  which IS the velocity.

and

The gradient on the v-t curve is  dv/dx which IS the acceleration.

-------------------------------------------------                                                    

log x^a = a*log x  so if a =1/5 then just multiply your number by 1/5.  In other words, just divide your number by 5.

 I'm sure you can do the division yourself.

Growth rate of juvenile is given by dh/dt = 0.049*(38 - h) where h is juvenile height and t is time.

This ordinary differential equation may be solved to find height, h, as a function of time, t, as follows:

Rewrite as dh/dt + 0.049h = 0.049*38

Multiply both sides by e^0.049t

e^0.049tdh/dt + e^0.049t*0.049h = 0.049*38*e^0.049t

Rewrite this as

d(e^0.049th)/dt = 0.049*38*e^0.049t

Integrate with respect to t

e^0.049th = 0.049*38*e^0.049t/0.049 + c    where c is a constant

Simplify:

e^0.049th = 38*e^0.049t + c

When t = 0 then h = 0 (i.e. zero height at birth) so

0 = 38 + c   or  c = -38

Therefore

e^0.049th = 38*e^0.049t - 38

Multiply both sides by e^-0.049t

h = 38(1 - e^-0.049t)

1. Plug in 94 for t on the right-hand side to find height at 94years.

2. Plug in 0.99*38 for h on the left-hand side and solve the resulting equation for t to find the age at colour-change.

Hi Dragon,

No Alan's is factorised - NOT simplified.

You cannot simplify this expression.     

2 1/4 = 9/4 = 2,25

The radioactive decay equation is N = N₀e^-ln(2)t/τ where N is number of atoms (or mass in this case), N0 is initial number (or mass) t is time and τ is half-life  (ln is the natural logarithm, or log to the base e).  So:

Mass after 5300 years = 626*e^{-ln(2)*5300/5730} grams.  I'll leave you to crunch the numbers.

Melody, I think your graphs are mainly ok.  I think I might draw the start of the accelerration curve with either a smaller initial value or a zero intial value, and then have a smoother change to a peak before it falls to point P.  However, this depends very much on the actual functional form of the x-t curve, which we don't know.  It's difficult to be sure from a sketch.

xvxvxv, at S the velocity is zero (see Melody's sketches).  Just to the left of S at time t1, say, the velocity is negative, just to the right, at time t2, it is positive. Acceleration is rate of change of velocity, so going from t1 to t2 the acceleration is (positive v - negative v)/(t2-t1).  Now, positive - negative is positive, and t2-t1 is positive, so overall the acceleration at S is positive. 

The mass is moving up the slope with an initial non-zero velocity.  The net-force on the mass is down the slope, all the way through the motion.

If velocity is taken as positive in the positive x-direction then for the uphill part of the motion velocity is going from, say, v1=-2m/s at time t1=0s to v2=0 m/s at time t2 = 1s (say).  The acceleration is therefore (v2-v1)/(t2-t1) = (0 - (-2))/(1 - 0) = +2m/s².

We don't intuit acceleration very well!!