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Long ago???...heck...at my age, yesterday is a blur....!!!!

Your Welcome Anonymous!

Most people forget what they had studied long ago and i too dont remember most of the things i learnt in like 2nd or 1st grade!But its all okay!

If you need any more help you are most welcome to ask!

thank you! i used to be the king of bedmas...but that was back in grade 8--over 12 years ago. i had 2 answers. 1 and 110

About 8 minutes....

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Cerca de 8 minutos....

Thank you CPhill!

Nice one, rosala.....I knew it would be easy for you....!!!!

Thumbs-Up and 3 points from me......

So we'd solve this lik

10x10+10(3-2)  ( now we will first multiply 10 by 3 and 2)

10x10+30-20  ( as according to DMAS ( Division Multiplication Addition Subtraction), we are going to do multiplication first , addition second and subtraction third as it says in the rule)

=100+30-20

=130-20

=110  ANSWER 

i hope uve understood by now but if not then pls be free to ask!

hope this helps and ive passed the test i hope!

Perimeter is given by....

P = 2(W + L)    so we have.....

18 = 2 (W + L )  divide by 2

9 = W + L     subtract W from both sides

9 - W  = L

Now...the area "formula" is given by

A = W * L    and we have

20 = W * L      ans substituting (9 - W) for L, we have

20 = W * (9 - W)

20 = -W² + 9W   .....Rearranging this gives

W² - 9W + 20 = 0

Factoring , we have

(W - 5) (W - 4)  = 0

And settong each factor to 0, we have either W = 4 or W = 5.....(depending on how we "orient" the rectangle.....)

We have.....

  lim          [ 6/(x+h) - 6/(x) ]   / h    =      

(h→0)

6*lim          [ 1/(x+h) - 1/(x) ]   / h    =    

   (h→0)

6* lim  [ (x) - (x+h)] / [(h)*(x)*(x + h)]  =

    (h→0)

6 * lim [-h] / [(h)*(x)*(x + h)]   =

     (h→0)

-6 * lim [ 1 / ( x² + xh) ]  =

      (h→0)

Now, let (h→0) and we have......

-6/x²   or    -6x^-2

Very nice, Alan....!!!

The employment of the Taylor series is a new one for me....(definitely a "tool chest" item!!)....When I was in school, I always wondered what the Taylor and McLaurin series were actually used for (other than to express something in terms of a polynomial!!)....I'm going to recommend this one as part of Melody's "End of the Day" Highlighted Problems...

Thumbs-Up from me....!!!

Let's take these inequalities one at a time..

x + 1 < 4     subtract 1 from both sides

x < 3

And the second inequality is

x - 8 > -7  add both 8 to both sides

x > 1

The  "∩" means "intersection"....we are asking...."What interval makes both inequalities true??"

Notice that the interval that makes both true is  1 < x < 3  which we can also denote as...(1, 3)

8^-8= ? *8^2    ....divide both sides by 8^2

(8^-8) /(8^2)  =  

(8^-8) * (8^-2)  =

8^-10

And that's the missing factor....

A. The least of the ... If the two sequences are continued,; what is the next number that will be seen in both sequences? Answer: ... Which set of shapes is arranged in the same pattern? A. B. C. D. ... Q07 Solve for W. R09 False algebraic expression.

Hey Melody

I know that CPhill is AMAZING and he always does these questions for me...so i myself, was taken aback when she said they were wrong. i will post how SHE does them for ya.

Here's the graph......

It's about 69.28° ...this could also be a second quadrant angle (since the sine is also positive there), i.e., ......

(180 - 69.28)  = about 110.72°

All of these are similar...I'll do the first one and give you a hint on two more....

(-3,2) (-1,1)

The idea is to find the slope (if it's not given) and then use something called the "point-slope formula" to write the equation in the "slope-intercept form."  (We could also write the equation in "standard form,' but I prefer slope-intercept unless somebody tells me different...!!)

OK.....the slope is defined as  (change in y) / (change in x)

So, from point 1 to point 2, y changes by -1. And from point 1 to point 2, x changes by +2.

So...our slope is -1/+2  = -(1/2)

Now...we'll use one of the points (doesn't matter which) to write an equation in point-slope form. I'll choose (-1,1).

The point-slope form is given by

y - y1 = m(x - x1)     ...where m is the slope, y1 is the y coordinate of our point and x1 is the x coordinate of our point....so we have......

y - 1 = -(1/2)(x - (-1))   =

y - 1  = -1/2(x) - 1/2     ....... add 1 to both sides.....

y = -1/2(x) + 1/2

And this is in slope-intercept form...(the y intercept is the "+1/2" at the end...)

The second one is the same...and the third one has the slope already given!!

Hint......for 4 ....the y-intercept = (0, -1)

And for 5, the x intercept is (6,0)   ........!!!!

Snell's law says that:  sin(θ₁)/sin(θ₂) = n₂/n₁ where the n's are the indices of refraction.  If the angle of incidence is zero, then the angle of refraction will be zero also, as the angles refer to the deviation from the normal to the surface.  However, I guess in your case you mean the angle from the surface, so that you are talking about grazing incidence.  In this case θ₁ is actually 90° and sin(θ₂) =  sin(90°)*1/1.5 = 1/1.5 so

$${\mathtt{theta2}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{1.5}}}}\right)} \Rightarrow {\mathtt{theta2}} = {\mathtt{41.810\: \!314\: \!895\: \!779^{\circ}}}$$

or θ₂ ≈ 41.81° (Note that this is also the angle made with the normal to the surface.  The angle made with the surface will be 90° - 41.81°.)

WolramAlpha gives......

And, of course, since it's an indefinite integal, we need to add a constant  !!!

P.S. - Someone else on the Forum might see a straightforward way to evaluate this, but it doesn't seem too easy, judging by the answer.......!!!

Hi Melody, Thats really nice that you are writing wraps daily! really like reading your wraps!Whenever i see that a wrap has been written on the forum i really get excited as if theres some treasure!its really enjoyable to read your wraps!

Thank you very much for them!

And those threads which appear below writing a question , they were at that place from long before!i had seen them a long ago!Andre is really nice , he does his work so quietly that no one even knows what has been done and just take the advantage of the things!

Thank you very much Andre for all the improvement!

I have obsereved the past few days that the message icon problem exist no more(at least with me , im sure with u all too), thank you very much for all the develpoment!

Heres something for you!

Hope you like it!!!

CAN ANYONE WRITE THE SOLUTION . THANKS

Corrected answer.......(thanks for catching my error, alan!!!)

Subtract 5*10^-10 from both sides

9 -5*10^-10 = log(x/5)      evaluate the left side

8.9999999995 = log(x/5)    

Exponentially, this says that

10^8.9999999995 = x/5          multiply both sides by 5

5*10^8.9999999995 = x  ≈ 4.9999999942435373 X 10⁹

Thurs 24/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

An explanations of some of my attitudes.

@@ End of Day Wrap :  Wed 23/7/14      Sydney, Australia     Time 11:50 pm      ♬                   

Hi all,

Great answers today were forthcoming from Heureka, Alan, Rosala, TakahiroMaeda, DragonSlayer554, CPhill, SquareRoot  and NinjaDevo.  Thanks to each of you.    

Now when a new thread is begun some "information for new users" displays underneath.   Admin has included 4 of our sticky threads in here.    

1) becomea member: why would you want to?

2) How do I type in fractions

3) Info for new people, and

4) LaTex.

I am not sure when Andre Massow added this feature.  I know it has been there for a couple of days at least.  Thank you Andre, we asked for something like this and true to form you have delivered.  

I only kept aside a couple of posts today,  here they are;

Unusual Trig (Thanks Heureka and T-rex)

Yes probably.