All Answers 358090 Answers

I'll help with a few, but I'm not going to give you the answers...I'll show you how to arrive at the answer for yourself.......deal???

1) When we have the form, y= mx ± b, the y intercept is given by  (0, ±b). Thus, if I have the equation y = -5x - 4, the y intercept is  just   (0, -4)...that should help you get started.....

2) To find the slope, subtract the y coordinates in any order. Then subtract the x coordinates in the same order. Then, put the first result over the second....that's the slope  !!!!    For example, suppose we have  (3,1) and (6,9). Subtract the y coordinates (9 - 1 = 8). Then subtract the x coordinates in the same order (6 - 3  = 3). Then the slope is 8/3.......

3) In equalities, I like to get the x's on the left side and everything else on the right. So. if I had, 5x - 5 <  2x + 9, I would first subtract the 2x from both sides giving 3x - 5 < 9. Now, add 5 to both sides and we have 3x < 14. Now divide both sides by 3 to get "x" on the left side by itself, and we have.......x < 14/3........

Work on these first and tell me how you're doing in a post.....I'd rather answer any questions you might have about these than going any further......

Hi Md,

That really is too many questions to put in all at once.

You have a go at them and if you really can't do them enter them one question per post onto the forum.

Try and tell us why you are having problems with them too or show us what you have attempted.

We are here to to help you learn, not to do all you home work for you.

Thank you

Melody

Thankyou Rosala,

You have a wonderful time.  We will miss you.

I liked the monkey one the best untill I saw the pest control guy.

That spider pest control guy sure knows how to take care of aracnids !!!!!!

Yes!Thanks!It will be lots of fun!Becoz it would be like a get together for me and all my cousins as we will all be meeting after a very long time!

Mon 28/7/14

Bye Bye from Roasala - Have fun, we'll see you in a week or so.   

Sounds like fun.......enjoy..............!!!!

@@ End of Day Wrap :     Sun 27/7/14     Sydney, Australia     Time 1:50am    (Really Monday morning)   ♬

Hi everyone,        ♫♪

Today our illustrious answerers were CPhill, Alan, Rosala and AzizHusain.  Thanks to each of you. 

Not a lot happened in Camelot today.                                                   ♫♪

Here are some interest threads but I don't think that any were outstanding.

A bit of fun                                 ♫♪

Yes CPhill!Im going to my grandmothers place for a festivals celebration thats on 29th July!

Are you going on a vacation???

"real number" was not specified either!

Perhap anonymous wanted to know how many sun glasses came after a trillion.

Answer.  A lot becasuse Sir CPhill gets them for free and he leaves them lying around EVERYWHERE!

I bet there are hundreds in those cupboards!

And Sir CPhill - there is no point arguing with me because I am a woman and therefore you have less than 5% chance of winning! And I am Queen of Camelot so you have no chance of winning.

And here are some more of those glasses that you keep losing!  

Please try to take better care of them!

Lady Guinevere.

Queen of Camelot

Otherwise known as the dinosaur!

True...but "integer" wasn't specified..........????

I think my sunglasses might be hiding in one of those cupboards......!!!!

Yes but you cannot find another integer in between a trillion and a trillion and one no matter what cupboard you look in.

Can you CPhill?        And I've got your sunnies!

Hi Sasini,

$$sin[\frac{\pi}{4}+A]*sin[\frac{\pi}{4}-A]\\\\
=[sin\frac{\pi}{4}cosA+cos\frac{\pi}{4}sinA][sin\frac{\pi}{4}cosA-cos\frac{\pi}{4}sinA]\\\\
\mbox{The two brackets are the same except one has a minus and the other a plus so this simplifies to the difference of 2 squares}\\\\
=[sin\frac{\pi}{4}cosA]^2-[cos\frac{\pi}{4}sinA]^2\\\\
=[sin^2\frac{\pi}{4}*cos^2A]-[cos^2\frac{\pi}{4}*sin^2A]\\\\
=[\frac{1}{2}cos^2A]-[\frac{1}{2}sin^2A]\\\\
=\frac{cos^2A-sin^2A}{2}\\\\
or\qquad
=\frac{1-2sin^2A}{2}\\\\
or\qquad
=\frac{cos2A}{2}$$

I got the same as well 

Using the sum and difference identities for the sine, we have......

[sin(pi/4)cosA + sin(A) cos(pi/4)] * [sin(pi/4)cosA - sinAcos(pi/4)] =

[sin(pi/4)cosA]² - (sin(A)cos(pi/4)]² =

[(1/2)cos²A] - [(1/2)sin²(A)] =

(1/2)[cos²A - sin²A] =

(1/2)cos2A

Which is the same as your answer, Sasini .....  !!!!

We can't say......no matter what you might say comes next, I can always find a real number that lies between your number and a trillion.....

However, if you're just interested in the names of "large" numbers, here you go:

(((4x-)^6y)^5)^2)

Puttin a minus sign there does not make sense Sasini.  

This is a very different question.

It is:

(((4x-)^6y)^5)^2)

=((4x)^6y)^10

=(4x)^60y

This is the answer for ur question

Firstly standard form is another name for scientific notation.

It is easiest to conver everything to metres first

0.03*0.055*0.04

$${\mathtt{0.03}}{\mathtt{\,\times\,}}{\mathtt{0.055}}{\mathtt{\,\times\,}}{\mathtt{0.04}} = {\mathtt{0.000\: \!066}}$$

= 6.6 x 10^-5  m³

It is 6.6 because you have to have one non-zero digit  in front of the decimal point.

It is 5 because I moved the decimal point 5 places.  It is negative because it is a very little number. 

(3^2)^2

=(3^2 =3 *3)^2

=(9)^2=9*9

=81

I think dr you can understand it. good luck!!

A trillion and one  

Aziz is correct 

I prefer to think of it as multiplying both sides by (x-2) and multiplying both sides by x

Equation must stay balanced.  What you do to one side you must do to the other side as well.

okay so now you have an answer for 69 and 96 lol

thanx for all 

Thanks Aziz for that great explanation.

I just thought i would show how the cancelling is done in fraction form.

(I am playing with LaTex,  some of this is a bit new for me  )

$$\dfrac{9x^5y^4}{4y}\times\dfrac{2y}{3x}\\\\\\
=\dfrac{\not{9} \textcolor[rgb]{0,0,1}{^3} x^{\not5 \textcolor[rgb]{0,0,1}{4}} y^4}{\not{4}\textcolor[rgb]{0,0,1}{^2}\not{y}}\times\dfrac{\not{2}\not{y}}{\not{3}\not{x}}\\\\\\\
=\dfrac{3x^4y^4}{2}$$

Melody is correct....there's not much you can do here, unless you would like to set this = 0 and graph it...in that case we get a "rotated conic"  (a hyperbola) which is kinda' cool.....(the presence of the "xy" term implies a "rotated" graph)