Thanks Chris :)

#PrayForBabaLlama I hope you feel better! 

yea i was tired part 3 will be longer

Also in this question, why they didn't wrote $$\frac{1}{2}(x+1)$$  insted of $$\frac{1}{2}x$$ ! 

Momentum is a vector quantity.  In the absence of any losses, the momentum of the system after the collision is the same as that before the collision. So, taking north as positive the total momentum of the system is:

1300*30 - 900*15 =  25500 kg.m/s

.

After the collision the velocity of the two cars, assuming they are stuck together, is

v = 25500/(1300 + 900)

or v = 11.59 m/s  (positive, so going north)

.

.

I think it must be sin4(x-3) insted of cos4(x-3) in the last step !

Yes!

Because: $$\int{\cos(4u)}\ du = \frac{1}{4}\sin(4u)$$

Two cars collide with each other. Before the collision, one car (m = 1300 kg) is going north at 30 m/s and the other car (m = 900 kg) is going south at 15 m/s. What is the momentum of the system made up of the two cars after the collision ?

$$\small{\text{
$
\boxed{v'_1 = v'_2 = \dfrac {(m_1\cdot v_1 - m_2\cdot v_2 )}{ (m_1+ m_2 ) } } \quad
\begin{array}{rcl}
m_1 &=& 1300\ \mathrm{kg} \\
v_1 &=& 30\ \frac{\mathrm{m} }{ \mathrm{s} }
\end{array}\quad
\begin{array}{rcl}
m_2 &=& 900\ \mathrm{kg} \\
v_2 &=& 15\ \frac{\mathrm{m} }{\mathrm{s}}
\end{array}
$
}}\\\\\\
\small{\text{
$
v'_1 = v'_2 = \dfrac {(1300\cdot 30 - 900\cdot 15 )}{ (1300+ 900 ) } \ \dfrac{\mathrm{m} }{\mathrm{s}}
$
}}\\\\
\small{\text{
$
v'_1 = v'_2 = \dfrac { 25500 }{ 2200 } \ \dfrac{ \mathrm{m}}{\mathrm{s}}
= 11.5909090909\ \dfrac{ \mathrm{m} }{\mathrm{s} }
$
}}\\\\
\small{\text{
$
m_1\cdot v'_1 + m_2\cdot v'_2 = 1300 * 11.591 + 900 * 11.591 = 25500\ \frac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}
$
}}$$

i dont hate school..its just tht i dont like it much!

but sometimes school isnt that bad...when you'll be old sitting couging on a rocking chair..you'll say "oh what days were those...when school used to entertain me *cough**cough*"

Never..as i pray!

and thanks CPhill!and same to you!

i forgot to mention...anyone who posts first on this thred gets 5 points!

there you go CPhill!(as expected)

Yes part 2 was a bit short

I do hope Bonnie is alright!

Find the radius of a sphere

$$\boxed{V=\dfrac{4}{3}\pi r^3 \qquad A=4\pi r^2}\\\\
r=\sqrt[3]{\dfrac{3V}{4\pi}} = \dfrac{1}{2}\sqrt{\dfrac{A}{\pi}}$$

Right triangle. Angle c= 90 degrees. Side A=5. Side B=15. Find angle b

$$\tan{(B)}}=\dfrac{15}{5}=3\\\\
B=\tan^{-1}{(3)}=71.5650511771\ensurement{^{\circ}}$$

You can

sin-1 is the same as asin

$$\dfrac{BK}{KD}
=\dfrac {1} { \dfrac{a}{b} }\cdot \left(1+\dfrac{d}{c} \right) \\\\\\
\dfrac{a}{b} = \dfrac{1}{2} \qquad \dfrac{d}{c} = \dfrac{2}{1} \\\\
\dfrac{BK}{KD} =
2\cdot \left( 1+2 \right) = 6 \\\\
\boxed{\dfrac{BK}{KD} = \dfrac{6}{1}}$$

Suppose you wish to solve: x-7 = 2x+15

You would plot the graph: y = x-7

and also plot: y = 2x+15

That really does look VERY impressive Heureka.

Hopefully I will have time to examine it properly later, because I don't have it now.  

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Hi anon,  Please read this post.

How to upload an Image.

-59=-20+13d    add 20 to each side

-39 = 13d         divide both sides by 13

-3 = d

Have a look at this graph, Melody.

(this is Mama)

Thanks CPhill! BabaLlama just fell asleep, so I'll tell him that he's in your prayers when he wakes up.

Llama on!

The probability of rolling a 6 would be 1/6 because there are 6 numbers but one 6.

The theoretical probability is 1/6 on any roll.

In 100 rolls, about 16 times.....more or less..

Me either!

Prayers, MamaLlama....!!!!

Can't wait for Part 3....!!!