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WE need more information. :)

sin(25)+sin(2)-cos(8)   in degrees

sin(25)+sin(2)-cos(8) = -0.532750310298

What is 'Les Miserables' about Chris?

Hi MJ

I read "Grapes of Wrath'  by John Steinbeck  fairly recently.

I also thought it was very good too!

It is not to old, I mean you could read and understand it alright but the content may be a little old for you to fully appreciate.  I don't think you should read the abridge version of this book.  It is not heavily padded with unnecessary content.  It should be read as is.  That is what I think.

Anne of Green Gables sounds like and excellent choice. 

Here is a serise of books that I think you might really enjoy.

They are light and funny stories about the adventures of a country vet in England :)

Please just put one question per post. :)

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

\(\large \frac{(a-1)^4-(b+1)^4}{(a-1)^3-(b+1)^3}=\frac{[(a-1)^2+(b+1)^2]\times [(a-1)^2-(b+1)^2]}{[(a-1)+(b+1)]\times [(a-1)-(b+1)]}\)

\(\large= \frac{(a^2-2a+1+b^2+2b+1)\times(a^2-2a+1-b^2-2b-1)}{(a+b)\times(a-b-2)}\)

\(\large= \frac{(a^2+2(b-a+1)+b^2)\times (a^2-2(a+b)-b^2)}{a^2-ab-2a+ab-b^2-2b}\)

\(\large= \frac{(a^2+2(b-a+1)+b^2)\times (a^2-2(a+b)-b^2)}{a^2-2(a+b)-b^2} \)

\(\large =a^2+2(b-a+1)+b^2\)    \(=a^2+2b-2a+2+b^2\)

  !

FV=P{[1 + R]^N - 1/ R}

5,707 = P{[1 +0.08]^25 -1 / 0.08}

5,707 = P x    73.10594...

PMT =5,707 / 73.10594...

PMT =$78.06

Constant acceleration kinetic equations can be used, as follows:

.

Can you expand on this question please?

If you double acceleration that means acceleration time 2.  

Where does you confusion spring from?

 6.3432452211234343213264888267457657692758777564967847310210023243565543674675444343232433223756890398109209280329783465439752309675203730828703623 percent%

[(x/2)2-(x/2)+k]+[(-x/2)2-(-x/2)+k]

Oh okay, thanks

I don't know, hectictar.....I'm not really a "Physics" person....

BTW....to be technically correct.....the answer to (3)  should be - 23.4 m/s

 Oh okay mine is wrong because I found the displacement not the distance, so when I tried to convert that to air time it just gave me something that was meaningless. I didn't really know what I was doing, heh.

But I'm confused though because when I tried to find the arc length from 0 to 90/49 I got about 8.61142 and then 18 divided by that is about 2.09. Howcome 2.09 seconds isn't the answer?

3. A 4kg object is 30m above the ground. 

If it falls what is the object velocity 2 meters above the ground? 

The time it takes to fall 28m  is :

-28  = -(1/2)9.81 * t^2

56/9.8  = t^2

sqrt (56/9.8)  = t ≈  2.39 s

And the velocity  is given by :

v(2.39)  =  9.8m/s^2 (2.39 s)   ≈ 23.4 m/s

2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air? 

We are using :

t  = 2 v₀ sin θ / g 

t =  2* 18m/s * sin (30)  / [9/8m /s^2]  =

36 * (1/2)  / 9.8    ≈  1.837 s

Here's a graph : https://www.desmos.com/calculator/2i436tqxnr

I also want to attempt the second one because why not haha I have no clue if I'm doing these right though.

So I'm going to work towards getting a position function of the ball.

Acceleration:

y''= -9.8 m/s²

Velocity:

y'=\(\int -9.8 dx\)

y' = -9.8x + c

18sin30° = -9.8(0) + c

c=9

y'= -9.8x + 9

Position:

y=\(\int(-9.8x+9)dx\)

y=\(\int-9.8x dx +\int9dx\)

y=\(-9.8\frac{x^{2}}{2}+9x\)

y= -4.9x² + 9x

So that is the position function of the ball. So we need to find the two places where the ball touches the ground, aka where the y value equals 0.

0 = -4.9x² + 9x

0 = x(-4.9x+9)

Set each factor equal to zero.

-4.9x+9 = 0 and x = 0

-4.9x=-9

x = 90/49

So y is 0 when x is 0 and when x is 90/49. The difference in those x values will be the distance it  traveled. So it traveled 90/49 meters. I just realized that we want the time it is in the air, not the distance traveled. Well it's okay the time it was in the air will just be 18 m/s divided by 90/49 meters. Which is 49/5 or 9.8 seconds. That was probably the longest possible approach to this problem but I kind of think I got the right answer.

Thanks.

My boo was a little different.  I did intend  0.32*6 and i knew the anwer was cents but then i wrote it wrong.

I am pleased that you picked the error up :)

uhm there isn't a button that looks like that in this calculator.

For the first one, I'm not really sure but I think that this is how to do it.

Imagine a graph of the velocity of the object.

Its velocity is 4 m/s at x = 0, and its velocity is 0 m/s at x=12 meters.

So if y is the velocity and x is the distance, the equation of that line is: y = (-4/12)x + 4, or

y = (-1/3)x + 4.

The derivative of velocity is acceleration, so the acceleration is -(1/3) m/s/s.

You know the acceleration is going to be negative because the object is slowing down.

I'm not sure if this is right though because I didn't use the information about it being 2 kg at all.

- [(+11.1)2 + (-6.1)2 + (-5.25) 2] ½  =

- [ 11.1 - 6.1 - 5.25 ]   =

- 11.1 + 6.1 + 5.25  =  .25   =  1/4

Unless this is supposed to be ;

- [ (11.1)^2 + (-6.1)^2 + (-5.25)^2] ^(1/2)  ≈ -13.7107

A little boo-boo.  0.32*0.06 =$0.0192 or 1.92cents

a + b  = 110

a - b  =    16       add the equations

2a  = 126      divide by 2

a  =  63    and

a - b  = 16

63 - b  = 16

63 - 16  = b   =   47

If the slope is  3.1,  this equals  3.1 / 1

So......everytime x changes by 1, y changes by 3.1

And x changes by 5  from  f(5) to f(10).....so....y changes by  5(3.1)  = 15.5

And adding that to its original value of 2 ....[ remember....(5.2)  is the original point ] we have

15.5 + 2  = 17.5