All Answers 358115 Answers

Remember that D/R = T. And since they travel the same time, let x be the distance that the car traveled and let 780-x  be the distance that the bus traveled. And the distance that the car traveled divided by its rate = the distance the bus traveled divided by its rate. So we have.....

x/68 = (780-x)/62

62x = 68(780-x)

62x = 53040 - 68x

130x = 53040

x = 408km

And this is the distance that the car traveled. And (780-408)km = 372km = the distance the bus traveled.

$${\frac{{\mathtt{9.45}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{15\,278.36}} = {\mathtt{1\,443.805\: \!02}}$$

Q-1) The answer is (A). Remember to subtract the length (1cm from the top and 1 cm from the bottom) already included in the other measurements.

Q-2 through Q-5 ) Answers are correct.

For Q-23 . . . . .

Your answer, (B), is (probably) correct. If it didn’t say “always” and/or have (D) “Any value” as an alternative, then I might be more certain.

This is an ambiguous question. (These are often found on physics tests).

For most equations friction work is (usually) signed as negative. Applied forces, are usually positive.

Technically, friction opposes motion and, hence, does negative work. However, friction is not a vector, so the negative sign does not tell us that it's moving backward.

Friction removes energy from the system. That is why it is (usually) signed as negative. To overcome this, energy needs to be (continuously) added to the system.

Nota bene: Another consideration is friction can be (0) zero. If this is the case, then it is neither positive nor negative. This really is an ambiguous question

~~D~~

P.S. Now, that I have reread the question, it is clear it refers to the previous question (Q-20). Your answer, (B), is correct.

It does help to read the whole question.

3. Correct.

4. Correct.

2. Correct.

1. 

I assume we have

(5y)x(2y^3) =

10y^4

In a closed system there is no loss of energy. The potential energy is the energy every object has that can be released by letting it fall, the potential energy will be transformed to kinetic energy (speed).

This means all the energy released by dropping 3 meters (5-2) will go towards kinetic energy. The formula for potential energy is E(potential)=mass*gravitational unit*height, the formula for kinetic energy is E(kinetic)=1/2*mass*velocity^2.

Since E(potential) becomes E(kenetic) you can equate these two kinds of energy, so E(potential)=E(kinetic) (more precisely E(potential_begin)-E(potential_end)=E(kenetic)). The mass is unknown but appears in both formulas so you remove it from both sides, the result is: height (difference in height)*gravitational unit=1/2*velocity^2 which grants you the answer to your question.

$$\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{y}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y3}}\right) = {\mathtt{30}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}$$

The side of the cube is 8 in

And the suface area is given by 6(s^2) = 6(8^2) = 384 sq in.

Look at Melody's diagram, rosala.  Note that the pink cuboid measures 5cm by 5 cm on one outside face. And Melody has noted that if we put a similar one underneath it and stack two more similar ones beside these two, we will have a face of a cube that will be 10 cm wide and 10 cm "tall." Now, look at the other side of the cuboid. It measures 2cm x 5 cm. And if we stack two of these together we will have the correct height of 10 cm. And if we do this 5 times, the width of the cube "facing us" will be 10 and its height will be 10 !!!

So the cube will neasure 10 x 10 x 10......

Does that nake sense??  (This is a tough one to see !!! Melody solved it very nicely, using the prime factors of 10.)

Just for the heck of it, try this with another number, say 14, whose prime factors are 2 and 7 !!!!!  It will work the same way !!!!

its ok Melody!i just thought u were angry becoz of your words which i didnt find a joke but as usual words cant express true feelings!

i think i just made a mistake in my post!anyways from next time i wont repeat it!thanks for correcting me!

Thank you all.

Your Welcome Melody!

That emu was just to catch your attention very well along with a laugh!im glad it got an owner!lol!

Your Welcome Melody!

That emu was just to catch your attention very well along with a laugh!im glad it got an owner!lol!

Thank you very much MElody for that help!i just solved this question with a very easy technique!and this questions sounds super easy to me now which was first looking like a lost land to me!

Anyways thank u so much for the help!i'd give u soo many points for that!lol!

Melody this sounds a lil confusing!im still not getting it!

(2/3) x (9/10) =

(2/10) * (9/3) =

(1/5) * 3

And you can finish evaluating this one.....

Okay melody thank you and I will wait 

I want more explanation for question 24 

Thanks Chris mine was right. I was just too tired. I said 24% of 25 was not a full number but of course it is. Silly me, I'll probably fix it in the morning. :)

Melody, I think we can set this up this way.....

Let x be the total number of students tested

Then, .24x ignored the question, 10 failed to answer it correctly, and 1.5(.24x) answered it correctly. So we have..........

.24x + 10 + 1.5(.24x) = x

.24x + 10 + .36x  = x

10 + .60x = x

10 = .40x

x = 25  = the total number of students tested

As a check    .24(25) = 6  and 1.5[.24(25)] = 9   and 6 + 10 + 9  = 25

Very nice, heureka.......I have not heard of this before........this is an area definitely worthy of a "new" exploration for me....!!!!

Sunday 31/8/14       

* 1) Proove that sqrt5 is irrational.

@@ End of Day Wrap:    Sat 30/8/14        Sydney, Australia      Time 2:45am (Really Sunday morning)      ♬

 Hi everyone,

It's been 2 days since my last but as always there have been many questions and more than that many answers.  Our illustrious answerers have been AzizHusain, NinjaDevo, Alan, DragonSlayer554, reptyGT, Rosala and CPhill.  Thank you all.  

reptyGT is new.  Welcome to our Web2 forum.  We hope that you like it here.  

I ma really pleased to see that the type of answers that are being provided have changed a little.  There have been less full answers and more help to get started.  This includes being sent to other web sites to learn things.  There are many great free sites available and we should make use of them at every oportunity.

When we tell someone to use the site calc I think it would be a good idea to add that it is on the home page.

I have put aside a lot of interest posts.  It has been 2 days since my last wrap.  Anyway here they are:

1) Ninja presents his full answers beautifully - just look at this equation solution !!